coupling by using dp - algorithm

Given two strings, S1 & S2. given scoring scheme where gap penalty, mismatch score and match score.
Find the S1 which have a best match with S2.
My idea is to list all possible S1 and then match one by one with S2. List all possible S1 by using brute force. Then match each possible S1 with S2 by using dp.
Is there is any faster way to do so? or suggest any reference?

Using Wikipedia and a little bit of thinking one could code up something like this:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#ifndef MAX
#define MAX(a,b) ((a)>(b)?(a):(b))
#endif
#define MATCH_SCORE 2
#define MISMATCH_SCORE -1
#define GAP_SCORE 0
// Calculates the match score recursively.
long MatchScore(const char* p/* in: may contain 'x' */,
const char* s/* in: doesn't contain 'x' */)
{
long res;
if (*p && *s)
{
if ((*p == *s) ||
((*p == 'x') && (*s >= 'a') && (*s <= 'f')))
{
res = MatchScore(p + 1, s + 1) + MATCH_SCORE;
}
else
{
long s1 = MatchScore(p + 1, s + 1) + MISMATCH_SCORE;
long s2 = MatchScore(p, s + 1) + GAP_SCORE;
long s3 = MatchScore(p + 1, s) + GAP_SCORE;
res = MAX(s1, MAX(s2, s3));
}
}
else
{
res = GAP_SCORE * (long)(strlen(p) + strlen(s));
}
return res;
}
// Calculates the best matching string and the match score
// using dynamic programming, the score must be the same
// as returned by MatchScore().
void FindBestMatch(const char* p/* in: may contain 'x' */,
const char* s/* in: doesn't contain 'x' */,
long* score/* out: match score */,
char** match/* out: best matching string */)
{
size_t lp = strlen(p) + 1;
size_t ls = strlen(s) + 1;
size_t i, j;
long* table = (long*)malloc(lp * ls * sizeof(long));
for (i = 0; i < lp; i++)
table[0 * lp + i] = GAP_SCORE * i;
for (j = 0; j < ls; j++)
table[j * lp + 0] = GAP_SCORE * j;
for (j = 1; j < ls; j++)
{
for (i = 1; i < lp; i++)
{
if ((p[i-1] == s[j-1]) ||
((p[i-1] == 'x') && (s[j-1] >= 'a') && (s[j-1] <= 'f')))
{
table[j * lp + i] = table[(j-1) * lp + (i-1)] + MATCH_SCORE;
}
else
{
table[j * lp + i] =
MAX(table[(j-1) * lp + (i-1)] + MISMATCH_SCORE,
MAX(table[(j-1) * lp + i] + GAP_SCORE,
table[j * lp + (i-1)] + GAP_SCORE));
}
}
}
*score = table[lp * ls - 1];
// Now, trace back the score table and construct the best matching string
*match = (char*)malloc(lp);
(*match)[lp - 1] = '\0';
for (j = ls, i = lp; j || i;)
{
if ((p[i-1] == s[j-1]) ||
((p[i-1] == 'x') && (s[j-1] >= 'a') && (s[j-1] <= 'f')))
{
(*match)[i-1] = s[j-1];
j--;
i--;
}
else
{
if (table[(j-1) * lp + i] > table[j * lp + (i-1)])
{
j--;
}
else
{
(*match)[i-1] = p[i-1];
i--;
}
}
}
free(table);
}
int main(void)
{
const char* pattern = "acdxdcxecxf";
const char* str = "abdfdaaed";
long score;
char* match;
char* match2;
printf("pattern=\"%s\" str=\"%s\"\n", pattern, str);
FindBestMatch(pattern, str, &score, &match);
printf("score=%ld (recursive)\n", MatchScore(pattern, str));
printf("score=%ld best match=\"%s\"\n", score, match);
// Now repeat with the best match we've just found,
// the result must be the same
printf("\nRepeating with pattern=best match:\n\n");
printf("pattern=\"%s\" str=\"%s\"\n", match, str);
FindBestMatch(match, str, &score, &match2);
printf("score=%ld (recursive)\n", MatchScore(match, str));
printf("score=%ld best match=\"%s\"\n", score, match2);
free(match);
free(match2);
return 0;
}
Output:
pattern="acdxdcxecxf" str="abdfdaaed"
score=14 (recursive)
score=14 best match="acdfdcaecdf"
Repeating with pattern=best match:
pattern="acdfdcaecdf" str="abdfdaaed"
score=14 (recursive)
score=14 best match="acdfdcaecdf"
I wonder if there're any bugs (other than the apparent lack of error checking).

Related

Segmentation fault is on terminal. Its the code to convert string to binary tree

Node * create(Node * root, int I, int J, string str)
{
if (I == J) { root -> data =str[I]; root -> left=NULL; root -> right=NULL; }
int i = 0, j = 0, k = 0, l = 0;
//// to store the data of root
string val;
for (i = I; i < J; i++) {
if (str[i] == '(') break;
val.push_back(str[i]);
}
root -> data = stoi(val);
stack < char > st;
for (j = i; j < J; j++) {
if (str[j] == '(') st.push(str[j]);
else if (str[j] == ')') st.pop();
if (st.empty()) break;
}
for (l = j + 1; l < J; l++) {
if (str[l] == '(') st.push(str[l]);
else if (str[l] == ')') st.pop();
if (st.empty()) break;
}
k = j + 1;
if (j - i == 2) root -> left -> data = str[i + 1];
else
root -> left = create(root -> left, i + 1, j - 1, str);
if (l == k) root -> right=NULL;
else if (l - k == 2) root -> right -> data = str[k + 1];
else
root -> right = create(root -> right, k + 1, l - 1, str);
return root;
}
Node * treeFromString(string str){
Node * p = create(p, 0, str.size() - 1, str);
return p;
}
Here I have initialized variables i , j , k , l to track the left child and right child bracket in the string. I , J are the range of the node for a particular activation record of recursion.
I assume you parsed some expression, by your code I pre-implemented it.
I just implement build a tree for following expression:
expression: |<symbol>| <integer> '(' <expression> ')' '('<expression>')'
symbol : any C++ `char` single byte character.
integer : any C++ valid int type value.
#include <iostream>
#include <string>
#include <string_view> // C++17 std::string_view
#include <charconv> // C++17 std::from_chars.
#include <cassert>
//simulate Node class
struct Node
{
Node *left, *right;
int data;
};
//1. use string_view instead of triple I, J, std::string.
//2. First argument Node* root - needn't. It should be local variable.
Node * create(std::string_view str)
{
assert(!str.empty());
if (str.size() == 1)
{
Node* root = new Node; //-> YOU should allocate a memory for root.
root -> data =str[0];
root -> left=nullptr; // use nullptr instead of NULL for c++11 or later.
root -> right=nullptr;
return root; // exit needed!
}
Node* root = new Node;
root->left = nullptr;
root->right = nullptr;
root->data = 0;
//// to store the data of root
//2. THERE NEED take an integer until first '(' symbol.
{
std::size_t i = 0;
while (i < str.size() && str[i] != '(' )
++i;
// str[0 .. i) - interval is an integer.
int val = 0;
(void)std::from_chars(str.data(), str.data() + i, val); // FOR simplifity don't check validness of conversation.
str.remove_prefix(i);
root->data = val;
}
//3. SKIP balanced '(' and ')'
/*stack < char > st;
for (j = i; j < J; j++) {
if (str[j] == '(') st.push(str[j]);
else if (str[j] == ')') st.pop();
if (st.empty()) break;
}
* */
/** we can implement it another way */
assert(!str.empty() && str[0] == '(' );
std::string_view within_bracket;
{
int balanced_brackets = 0;
std::size_t i = 0;
while (i < str.size())
{
if (str[i] == '(') ++ balanced_brackets;
else if (str[i] == ')' ) --balanced_brackets;
i++;
if (balanced_brackets == 0)
break;
}
assert (i > 0 && str[ i - 1 ] == ')' );
// 0 1 2 3
// str[0..i) - is '(' ... ')' symbols.
within_bracket = std::string_view(str.data() + 1, i - 2);
str.remove_prefix(i);
}
/****4. THIS second balanced bracket check */
std::string_view second_within_bracket;
/*
for (l = j + 1; l < J; l++) {
if (str[l] == '(') st.push(str[l]);
else if (str[l] == ')') st.pop();
if (st.empty()) break;
}
k = j + 1;
*/
assert(!str.empty() && str[0] == '(' );
// ========== second balanced brackets check ==========
{
std::size_t i = 0;
int balanced_brackets = 0;
while (i < str.size())
{
if (str[i] == '(') ++ balanced_brackets;
else if (str[i] == ')' ) --balanced_brackets;
i++;
if (balanced_brackets == 0)
break;
}
// 0 1 2 3
// str[0..i) - is '(' ... ')' symbols.
second_within_bracket = std::string_view(str.data() + 1, i - 2);
str.remove_prefix(i);
}
//================================
/*
if (j - i == 2) root -> left -> data = str[i + 1];
else
root -> left = create(i + 1, j - 1, str);
if (l == k) root -> right=NULL;
else if (l - k == 2) root -> right -> data = str[k + 1];
else
root -> right = create(root -> right, k + 1, l - 1, str);
*/
root->left = create(within_bracket);
root->right = create(second_within_bracket);
return root;
}
Node * treeFromString(std::string_view str){
Node * p = create(str);
return p;
}
void printTree(Node* root, int level = 0)
{
if (root==nullptr) return;
for (int i= 0; i < level; ++i) std::cout << "--";
std::cout << " data = " << root->data << std::endl;
printTree(root->left, level + 1);
printTree(root->right, level + 1);
}
int main(){
std::string str = "12(8(7)(5))(9(3)(2(1)(8)))";
Node * expr = treeFromString(str);
printTree(expr);
}
godbold output:
Program returned: 0
data = 12
-- data = 8
---- data = 55
---- data = 53
-- data = 9
---- data = 51
---- data = 2
------ data = 49
------ data = 56
This answer is a bit different, it assumes you are loading a tree of integer values from a string, that could be loaded from a file. Next time you ask a question about code, could you please explain a little bit what the code does? Guessing does take some time and some effort.
I've reused the main() and PrintTree() functions from Khurshid Normuradov's answer. I hope he won't mind.
I took the liberty to add modern c++ coding techniques because this question is tagged c++17, so you're getting an example in c++17.
#include <algorithm>
#include <iostream>
#include <memory>
#include <string>
#include <string_view>
struct Node {
std::unique_ptr<Node> left = nullptr;
std::unique_ptr<Node> right = nullptr;
int value = 0;
};
std::unique_ptr<Node> treeFromString(std::string_view str) {
std::cout << "treeFromString(\"" << str << "\")\n";
if (str.empty()) return {};
auto node = std::make_unique<Node>();
// extract an int
auto pos = str.find_first_not_of("0123456789");
auto val = str.substr(0, pos);
// optional: std::stoi() would throw anyway in this case
if (val.empty())
throw std::runtime_error("invalid value in expression");
node->value = std::stoi(std::string{val});
if (val.length() == str.length()) return node;
str = str.substr(val.length());
// Both left/right parsing are similar and use this
// common subroutine.
// expects parens delimited string as input.
// param str in: str string to parse from
// out: whatever's left to parse
// returns string content within parens, parens not included.
auto extract_parens_contents = [](std::string_view& str) {
// right here would be the perfect place to insert code to skip
// whitespace if you ever needed to do that.
// find parens extent
int parens = 0;
auto parens_end =
std::find_if(str.begin(), str.end(), [&parens](char c) {
parens += (c == '(') - (c == ')');
return (parens == 0);
});
if (parens_end == str.end())
throw std::runtime_error("unbalanced parens in expression");
// extract result
auto result = std::string_view(
str.begin() + 1, std::distance(str.begin() + 1, parens_end));
// remove spent bytes from input stream
str = std::string_view(
parens_end + 1,
str.length() - std::distance(str.begin(), parens_end + 1));
return result;
};
node->left = treeFromString(extract_parens_contents(str));
node->right = treeFromString(extract_parens_contents(str));
return node;
}
// special thanks to user Khurshid Normuradov, who originally wrote the two functions below.
// it would be difficult to writing something that would be any better for the
// intended purpose.
void printTree(Node* root, int level = 0) {
if (root == nullptr) return;
for (int i = 0; i < level; ++i) std::cout << "--";
std::cout << " data = " << root->value << std::endl;
printTree(root->left.get(), level + 1);
printTree(root->right.get(), level + 1);
}
int main() {
std::string str = "12(8(7)(5))(9(3)(2(1)(8)))";
auto expr = treeFromString(str);
printTree(expr.get());
}
You can play with the code here: https://godbolt.org/z/ch3zv5KTT

Time limit exceeded in my code given below

Question:
Lapindrome is defined as a string which when split in the middle, gives two halves having the same characters and same frequency of each character. If there are odd number of characters in the string, we ignore the middle character and check for lapindrome. For example gaga is a lapindrome, since the two halves ga and ga have the same characters with same frequency. Also, abccab, rotor and xyzxy are a few examples of lapindromes. Note that abbaab is NOT a lapindrome. The two halves contain the same characters but their frequencies do not match.
Your task is simple. Given a string, you need to tell if it is a lapindrome.
Input:
First line of input contains a single integer T, the number of test cases.
Each test is a single line containing a string S composed of only lowercase English alphabet.
Output:
For each test case, output on a separate line: "YES" if the string is a lapindrome and "NO" if it is not.
Constraints:
1 ≤ T ≤ 100
2 ≤ |S| ≤ 1000, where |S| denotes the length of S
#include <stdio.h>
#include <string.h>
int found;
int lsearch(char a[], int l, int h, char p) {
int i = l;
for (i = l; i <= h; i++) {
if (a[i] == p) {
found = 0;
return i;
}
}
return -1;
}
int main() {
char s[100];
int q, z, i, T;
scanf("%d", &T);
while (T--) {
q = 0;
scanf("%s", &s);
if (strlen(s) % 2 == 0)
for (i = 0; i < (strlen(s) / 2); i++) {
z = lsearch(s, strlen(s) / 2, strlen(s) - 1, s[i]);
if (found == 0) {
found = -1;
s[z] = -2;
} else
q = 1;
} else
for (i = 0; i < (strlen(s) / 2); i++) {
z = lsearch(s, 1 + (strlen(s) / 2), strlen(s) - 1, s[i]);
if (found == 0) {
found = -1;
s[z] = -2;
} else
q = 1;
}
if (strlen(s) % 2 == 0)
for (i = (strlen(s) / 2); i < strlen(s); i++) {
if (s[i] != -2)
q = 1;
} else
for (i = (strlen(s) / 2) + 1; i < strlen(s); i++) {
if (s[i] != -2)
q = 1;
}
if (q == 1)
printf("NO\n");
else
printf("YES\n");
}
}
I am getting correct output in codeblocks but the codechef compiler says time limit exceeded. Please tell me why it says so
For each of O(n) characters you do a O(n) search leading to a O(n^2) algorithm. Throw a thousand character string at it, and it is too slow.
This is solvable in two standard ways. The first is to sort each half of the string and then compare. The second is to create hash tables for letter frequency and then compare.

Maximum product prefix string

The following is a demo question from a coding interview site called codility:
A prefix of a string S is any leading contiguous part of S. For example, "c" and "cod" are prefixes of the string "codility". For simplicity, we require prefixes to be non-empty.
The product of prefix P of string S is the number of occurrences of P multiplied by the length of P. More precisely, if prefix P consists of K characters and P occurs exactly T times in S, then the product equals K * T.
For example, S = "abababa" has the following prefixes:
"a", whose product equals 1 * 4 = 4,
"ab", whose product equals 2 * 3 = 6,
"aba", whose product equals 3 * 3 = 9,
"abab", whose product equals 4 * 2 = 8,
"ababa", whose product equals 5 * 2 = 10,
"ababab", whose product equals 6 * 1 = 6,
"abababa", whose product equals 7 * 1 = 7.
The longest prefix is identical to the original string. The goal is to choose such a prefix as maximizes the value of the product. In above example the maximal product is 10.
Below is my poor solution in Java requiring O(N^2) time. It is apparently possible to do this in O(N). I was thinking Kadanes algorithm. But I can't think of any way that I can encode some information at each step that lets me find the running max. Can any one think of an O(N) algorithm for this?
import java.util.HashMap;
class Solution {
public int solution(String S) {
int N = S.length();
if(N<1 || N>300000){
System.out.println("Invalid length");
return(-1);
}
HashMap<String,Integer> prefixes = new HashMap<String,Integer>();
for(int i=0; i<N; i++){
String keystr = "";
for(int j=i; j>=0; j--) {
keystr += S.charAt(j);
if(!prefixes.containsKey(keystr))
prefixes.put(keystr,keystr.length());
else{
int newval = prefixes.get(keystr)+keystr.length();
if(newval > 1000000000)return 1000000000;
prefixes.put(keystr,newval);
}
}
}
int maax1 = 0;
for(int val : prefixes.values())
if(val>maax1)
maax1 = val;
return maax1;
}
}
Here's a O(n log n) version based on suffix arrays. There are O(n) construction algorithms for suffix arrays, I just don't have the patience to code them.
Example output (this output isn't O(n), but it's only to show that we can indeed compute all the scores):
4*1 a
3*3 aba
2*5 ababa
1*7 abababa
3*2 ab
2*4 abab
1*6 ababab
Basically you have to reverse the string, and compute the suffix array (SA) and the longest common prefix (LCP).
Then you have traverse the SA array backwards looking for LCPs that match the entire suffix (prefix in the original string). If there's a match, increment the counter, otherwise reset it to 1. Each suffix (prefix) receive a "score" (SCR) that corresponds to the number of times it appears in the original string.
#include <iostream>
#include <cstring>
#include <string>
#define MAX 10050
using namespace std;
int RA[MAX], tempRA[MAX];
int SA[MAX], tempSA[MAX];
int C[MAX];
int Phi[MAX], PLCP[MAX], LCP[MAX];
int SCR[MAX];
void suffix_sort(int n, int k) {
memset(C, 0, sizeof C);
for (int i = 0; i < n; i++)
C[i + k < n ? RA[i + k] : 0]++;
int sum = 0;
for (int i = 0; i < max(256, n); i++) {
int t = C[i];
C[i] = sum;
sum += t;
}
for (int i = 0; i < n; i++)
tempSA[C[SA[i] + k < n ? RA[SA[i] + k] : 0]++] = SA[i];
memcpy(SA, tempSA, n*sizeof(int));
}
void suffix_array(string &s) {
int n = s.size();
for (int i = 0; i < n; i++)
RA[i] = s[i] - 1;
for (int i = 0; i < n; i++)
SA[i] = i;
for (int k = 1; k < n; k *= 2) {
suffix_sort(n, k);
suffix_sort(n, 0);
int r = tempRA[SA[0]] = 0;
for (int i = 1; i < n; i++) {
int s1 = SA[i], s2 = SA[i-1];
bool equal = true;
equal &= RA[s1] == RA[s2];
equal &= RA[s1+k] == RA[s2+k];
tempRA[SA[i]] = equal ? r : ++r;
}
memcpy(RA, tempRA, n*sizeof(int));
}
}
void lcp(string &s) {
int n = s.size();
Phi[SA[0]] = -1;
for (int i = 1; i < n; i++)
Phi[SA[i]] = SA[i-1];
int L = 0;
for (int i = 0; i < n; i++) {
if (Phi[i] == -1) {
PLCP[i] = 0;
continue;
}
while (s[i + L] == s[Phi[i] + L])
L++;
PLCP[i] = L;
L = max(L-1, 0);
}
for (int i = 1; i < n; i++)
LCP[i] = PLCP[SA[i]];
}
void score(string &s) {
SCR[s.size()-1] = 1;
int sum = 1;
for (int i=s.size()-2; i>=0; i--) {
if (LCP[i+1] < s.size()-SA[i]-1) {
sum = 1;
} else {
sum++;
}
SCR[i] = sum;
}
}
int main() {
string s = "abababa";
s = string(s.rbegin(), s.rend()) +".";
suffix_array(s);
lcp(s);
score(s);
for(int i=0; i<s.size(); i++) {
string ns = s.substr(SA[i], s.size()-SA[i]-1);
ns = string(ns.rbegin(), ns.rend());
cout << SCR[i] << "*" << ns.size() << " " << ns << endl;
}
}
Most of this code (specially the suffix array and LCP implementations) I have been using for some years in contests. This version in special I adapted from this one I wrote some years ago.
public class Main {
public static void main(String[] args) {
String input = "abababa";
String prefix;
int product;
int maxProduct = 0;
for (int i = 1; i <= input.length(); i++) {
prefix = input.substring(0, i);
String substr;
int occurs = 0;
for (int j = prefix.length(); j <= input.length(); j++) {
substr = input.substring(0, j);
if (substr.endsWith(prefix))
occurs++;
}
product = occurs*prefix.length();
System.out.println("product of " + prefix + " = " +
prefix.length() + " * " + occurs +" = " + product);
maxProduct = (product > maxProduct)?product:maxProduct;
}
System.out.println("maxProduct = " + maxProduct);
}
}
I was working on this challenge for more than 4 days , reading a lot of documentation, I found a solution with O(N) .
I got 81%, the idea is simple using a window slide.
def solution(s: String): Int = {
var max = s.length // length of the string
var i, j = 1 // start with i=j=1 ( is the beginning of the slide and j the end of the slide )
val len = s.length // the length of the string
val count = Array.ofDim[Int](len) // to store intermediate results
while (i < len - 1 || j < len) {
if (i < len && s(0) != s(i)) {
while (i < len && s(0) != s(i)) { // if the begin of the slide is different from
// the first letter of the string skip it
i = i + 1
}
}
j = i + 1
var k = 1
while (j < len && s(j).equals(s(k))) { // check for equality and update the array count
if (count(k) == 0) {
count(k) = 1
}
count(k) = count(k) + 1
max = math.max((k + 1) * count(k), max)
k = k + 1
j = j + 1
}
i = i + 1
}
max // return the max
}

Run length encoding using O(1) space

Can we do the run-length encoding in place(assuming the input array is very large)
We can do for the cases such as AAAABBBBCCCCDDDD
A4B4C4D4
But how to do it for the case such as ABCDEFG?
where the output would be A1B1C1D1E1F1G1
My first thought was to start encoding from the end, so we will use the free space (if any), after that we can shift the encoded array to the start. A problem with this approach is that it will not work for AAAAB, because there is no free space (it's not needed for A4B1) and we will try to write AAAAB1 on the first iteration.
Below is corrected solution:
(let's assume the sequence is AAABBC)
encode all groups with two or more elements and leave the rest unchanged (this will not increase length of the array) -> A3_B2C
shift everything right eliminating empty spaces after first step -> _A3B2C
encode the array from the start (reusing the already encoded groups of course) -> A3B2C1
Every step is O(n) and as far as I can see only constant additional memory is needed.
Limitations:
Digits are not supported, but that anyway would create problems with decoding as Petar Petrov mentioned.
We need some kind of "empty" character, but this can be worked around by adding zeros: A03 instead of A3_
C++ solution O(n) time O(1) space
string runLengthEncode(string str)
{
int len = str.length();
int j=0,k=0,cnt=0;
for(int i=0;i<len;i++)
{
j=i;
cnt=1;
while(i<len-1 && str[i]==str[i+1])
{
i++;
cnt++;
}
str[k++]=str[j];
string temp =to_string(cnt);
for(auto m:temp)
str[k++] = m;
}
str.resize(k);
return str;
}
null is used to indicate which items are empty and will be ignored for encoding. Also you can't encode digits (AAA2222 => A324 => 324 times 'A', but it's A3;24). Your question opens more questions.
Here's a "solution" in C#
public static void Encode(string[] input)
{
var writeIndex = 0;
var i = 0;
while (i < input.Length)
{
var symbol = input[i];
if (symbol == null)
{
break;
}
var nextIndex = i + 1;
var offset = 0;
var count = CountSymbol(input, symbol, nextIndex) + 1;
if (count == 1)
{
ShiftRight(input, nextIndex);
offset++;
}
input[writeIndex++] = symbol;
input[writeIndex++] = count.ToString();
i += count + offset;
}
Array.Clear(input, writeIndex, input.Length - writeIndex);
}
private static void ShiftRight(string[] input, int nextIndex)
{
var count = CountSymbol(input, null, nextIndex, (a, b) => a != b);
Array.Copy(input, nextIndex, input, nextIndex + 1, count);
}
private static int CountSymbol(string[] input, string symbol, int nextIndex)
{
return CountSymbol(input, symbol, nextIndex, (a, b) => a == b);
}
private static int CountSymbol(string[] input, string symbol, int nextIndex, Func<string, string, bool> cmp)
{
var count = 0;
var i = nextIndex;
while (i < input.Length && cmp(input[i], symbol))
{
count++;
i++;
}
return count;
}
The 1st solution does not take care of single characters. For example - 'Hi!' will not work. I've used totally different approach, used 'insert()' functions to add inplace. This take care of everything, whether the total 'same' character is > 10 or >100 or = 1.
#include<iostream>
#include<algorithm>
using namespace std;
int main(){
string name = "Hello Buddy!!";
int start = 0;
char distinct = name[0];
for(int i=1;i<name.length()+1;){
if(distinct!=name[i]){
string s = to_string(i-start);
name.insert(start+1,s);
name.erase(name.begin() + start + 1 + s.length(),name.begin() + s.length() + i);
i=start+s.length()+1;
start=i;
distinct=name[start];
continue;
}
i++;
}
cout<<name;
}
Suggest me if you find anything incorrect.
O(n), in-place RLE, I couldn't think better than this. It will not place a number, if chars occurence is just 1. Will also place a9a2, if the character comes 11 times.
void RLE(char *str) {
int len = strlen(str);
int count = 1, j = 0;
for (int i = 0; i < len; i++){
if (str[i] == str[i + 1])
count++;
else {
int times = count / 9;
int rem = count % 9;
for (int k = 0; k < times; k++) {
str[j++] = str[i];
_itoa(9, &str[j++], 10);
count = count - 9;
}
if (count > 1) {
str[j++] = str[i];
_itoa(rem, &str[j++], 10);
count = 1;
}
else
str[j++] = str[i];
}
}
cout << str;
}
I/P => aaabcdeeeefghijklaaaaa
O/P => a3bcde4fghijkla5
Inplace solution using c++ ( assumes length of encoding string is not more than actual string length):
#include <bits/stdc++.h>
#include<stdlib.h>
using namespace std;
void replacePattern(char *str)
{
int len = strlen(str);
if (len == 0)
return;
int i = 1, j = 1;
int count;
// for each character
while (str[j])
{
count = 1;
while (str[j] == str[j-1])
{
j = j + 1;
count++;
}
while(count > 0) {
int rem = count%10;
str[i++] = to_string(rem)[0];
count = count/10;
}
// copy character at current position j
// to position i and increment i and j
if (str[j])
str[i++] = str[j++];
}
// add a null character to terminate string
if(str[len-1] != str[len-2]) {
str[i] = '1';
i++;
}
str[i] = '\0';
}
// Driver code
int main()
{
char str[] = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabccccc";
replacePattern(str);
cout << str;
return 0;
}

Dynamic Programming Altogorithm

I'm trying to construct an algorithm that runs at O(nb) time with the following input/question:
input: an array A[1..n] of n different integers and an integer b (i am assuming that the numbers in A are sequential, starting at 1 ending at n, i.e. for n=4 A[1,2,3,4].
question: in how many ways can b be written as the sum of elements of the array when elements in A[] can only be used once?
I've kind of hit a wall on this one. I'm looking for some kind of recursive solution, but I don't see how to avoid using repeat numbers. Like, for instance, if we started at 1 and stored all the ways to make one (just 1) then 2 (just 2) then three (3 or 2+1) etc, it shouldn't be hard to see how many ways we can make larger numbers. But if, for instance, we take 5, we will see that it can be broken into 4+1, and 4 can be further broken down into 3+1, so then we would see 2 solutions (4+1, and 3+1+1), but one of those has a repeat of a number. Am I missing something obvious? Thanks so much!
Recursive and dynamic solutions in C:
#include <stddef.h>
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
typedef unsigned char uchar;
typedef unsigned int uint;
typedef struct tAddend
{
struct tAddend* pPrev;
uint Value;
} tAddend;
void findRecursiveSolution(uint n, uint maxAddend, tAddend* pPrevAddend)
{
uint i;
for (i = maxAddend; ; i--)
{
if (n == 0)
{
while (pPrevAddend != NULL)
{
printf("+%u", pPrevAddend->Value);
pPrevAddend = pPrevAddend->pPrev;
}
printf("\n");
return;
}
if (n >= i && i > 0)
{
tAddend a;
a.pPrev = pPrevAddend;
a.Value = i;
findRecursiveSolution(n - i, i - 1, &a);
}
if (i <= 1)
{
break;
}
}
}
void printDynamicSolution(uchar** pTable, uint n, uint idx, uint sum, tAddend* pPrevAddend)
{
uchar el = pTable[idx][sum];
assert((el != 0) && (el != 5) && (el != 7));
if (el & 2) // 2,3,6 - other(s)
{
printDynamicSolution(pTable,
n,
idx - 1,
sum,
pPrevAddend);
}
if (el & 4) // self + other(s)
{
tAddend a;
a.pPrev = pPrevAddend;
a.Value = idx + 1;
printDynamicSolution(pTable,
n,
idx - 1,
sum - (idx + 1),
&a);
}
if (el & 1) // self, found a solution
{
tAddend a;
a.pPrev = pPrevAddend;
a.Value = idx + 1;
pPrevAddend = &a;
while (pPrevAddend != NULL)
{
printf("+%u", pPrevAddend->Value);
pPrevAddend = pPrevAddend->pPrev;
}
printf("\n");
}
}
void findDynamicSolution(uint n)
{
uchar** table;
uint i, j;
if (n == 0)
{
return;
}
// Allocate the DP table
table = malloc(sizeof(uchar*) * n);
if (table == NULL)
{
printf("not enough memory\n");
return;
}
for (i = 0; i < n; i++)
{
table[i] = malloc(n + 1);
if (table[i] == NULL)
{
while (i > 0)
{
free(table[--i]);
}
free(table);
printf("not enough memory\n");
return;
}
}
// Fill in the DP table
for (i = 0; i < n; i++)
{
for (j = 0; j <= n; j++)
{
if (i == 0)
{
table[i][j] = (i + 1 == j); // self
}
else
{
table[i][j] = (i + 1 == j) + // self
2 * (table[i - 1][j] != 0) + // other(s)
4 * ((j >= i + 1) && (table[i - 1][j - (i + 1)] != 0)); // self + other(s)
}
}
}
printDynamicSolution(table, n, n - 1, n, NULL);
for (i = 0; i < n; i++)
{
free(table[i]);
}
free(table);
}
int main(int argc, char** argv)
{
uint n;
if (argc != 2 || sscanf(argv[1], "%u", &n) != 1)
{
n = 10;
}
printf("Recursive Solution:\n");
findRecursiveSolution(n, n, NULL);
printf("\nDynamic Solution:\n");
findDynamicSolution(n);
return 0;
}
Output:
for 10:
Recursive Solution:
+10
+1+9
+2+8
+3+7
+1+2+7
+4+6
+1+3+6
+1+4+5
+2+3+5
+1+2+3+4
Dynamic Solution:
+1+2+3+4
+2+3+5
+1+4+5
+1+3+6
+4+6
+1+2+7
+3+7
+2+8
+1+9
+10
See also on ideone.
Let F(x,i) be the number of ways elements of A[1:i] can be summed to get x.
F(x,i+1) = F(x-A[i+1],i) + F(x,i)
That is it!
This is not a dynamic programming solution though. Non-recursive.
Assumption that arr is sorted in your case like [i....j] where a[i] <= a[j]
That's easy enough
void summer(int[] arr, int n , int b)
{
int lowerbound = 0;
int upperbound = n-1;
while (lowerbound < upperbound)
{
if(arr[lowerbound]+arr[upperbound] == b)
{
// print arr[lowerbound] and arr[upperbound]
lowerbound++; upperbound--;
}
else if(arr[lowerbound]+arr[upperbound] < b)
lowerbound++;
else
upperbound--;
}
}
The above program is easily modifiable to a recursive you need to only change the function definition by passing lowerbound and upperbound.
Case for termination is still lowerbound < upperbound
Base case is if arr[lowerbound] +arr[upperbound] == b
Edited based on comments
You will need to use a modified version of integer knapsack problem. The values of [i,j] both need to be modified accordingly. You are having the problem because you are not most probably modifying your i carefully, Increase your i accordingly then their will not be repetition like the one you are having.

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