finding minimum number in a integer - algorithm

Question

Given an n-digit Number and a number 'k'. You have to remove ‘k’ digits from the number and give the shortest number from the remaining ‘n-k’ digits such that the sequence of digits remains same. For example, if the number is 637824 and k = 3. So you have to remove 3 digits from the given number. The number formed from the remaining digits should be the smallest possible and the sequence of digits must not be changed. So the output should be 324.
Approach which i have used same as inclusion exclusion logic:
Input 63119 and K=2:
Select 9 + minimum of(6311) = 19 or
don't select 9, select 1 + minimum of(631) = 11 and finally take minimum of all.
For input 4567813 and k=3:
Select 3 and 1 along with minimum of(45678) = 413
I am using recursion logic to get this right but i am not able to implement this with code and i am spent force now. I need help with this recursion. I am not after better solution.
#define min(a, b) ((a)>(b)?(b):(a));
int minimum(char *s, int i, int j)
{
if (i == j)
return s[i] - '0';
return min(s[j]-'0', minimum(s, i, j-1));
}
int add_up(char *s, int i, int j)
{
int sum = 0, mul = 1;
while(i < j) {
sum = sum + (s[j] - '0')*mul;
j--;mul *= 10;
}
return sum;
}
int foo(char *s, int size, int j, int k)
{
int sum = 0, i, mul = 1;
if (k < 0 || j > size || j < 0)
return 0;
if ((k == 0) && (j != 0))
return add_up(s, 0, j);
if ((k == 1) && (j != 0))
return minimum(s, 0, j);
if (k-1 == j)
return add_up(s, 0, j);
for (i=k;i>=0;i--) {
sum += min((s[j]-'0')+10*foo(s, size, j-1, k-1), foo(s, size, j-1, k));
}
return sum;
}
int main()
{
char s[] = {"4567813"};
printf("%d\n", foo(s, strlen(s)-1, strlen(s)-1, 2));
return 0;
}

Answer 1

You have said finally take minimum of all but you have taken minimum of all and then added them. You need to take minimum for each i and save minimum over all i in sum. See the code for clarification. Also there is bug in your add_up function, it should add until i<=j.
int add_up(char *s, int i, int j)
{
int sum=0, mul=1;
while(i <= j) { // modification here
sum=sum + (s[j] - '0')*mul;
j--; mul*=10;
}
return sum;
}
int foo(char *s, int size, int j, int k)
{
int sum=INT_MAX, i, mul=1;
if(k < 0 || j > size || j < 0)
return 0;
if((k == 0) && (j != 0))
return add_up(s, 0, j);
if((k == 1) && (j != 0))
return minimum(s, 0, j);
if(k-1 == j)
return add_up(s, 0, j);
for(i=k; i>=0; i--) {
int res=min((s[j]-'0')+10*foo(s, size, j-1, k-1), foo(s, size, j-1, k));
sum=min(sum,res); // minimum over all possible i
}
return sum;
}