Count Operations to Reduce a Number in C++

We are given with a positive integer K and an array Ops[] which contains integers. The goal is to find the number of operations required to reduce K such that it becomes less than 0. Operations are −

• First operation is K + Ops[0], first element added to K

• After 1. Add Ops[i] to K until K<0. where index i keeps on changing in a circular manner. 0<=i<N. N is the number of integers in Ops[].

Note − Keep adding Ops[i] until K<0. If i reaches the last element Ops[N-1] then again start from i=0. In a circular manner.

We will first check if the sum of all elements of array Ops[] >0. If yes then K can never be reduced. Return -1. Otherwise keep adding Ops[i] to K and check if K<0 if yes break the loop.

Increment count of operations after addition: K+Ops[i].

Let’s understand with examples.

Input −

ops[]= { -4,2,-3,0,2 }, K=5

Output − Count of operations required to reduce a number − 3

Explanation − K is 5. Operations are −

1. K+ops[0]= 5+(-4) = 1
2. K+ops[1]= 1+2 = 3
3. K+ops[2]= 3+(-3) = 0

Input −

ops[]= { 5,5,3,-2 }, K=10

Output − K cannot be reduced!!

Explanation −K is 10. Operations are −

1. K+ops[0]= 10+5= 15
2. K+ops[1]= 15+5= 20
3. K+ops[2]= 20+3= 23
4. K+ops[3]= 23+-2= 22
5. K+ops[0]= 22+5= 27
6. K+ops[1]= 27+5=32
7. …………………

If we early check the sum of all elements of ops[]=5+5+3-2=11 and 11+10 is always +ve. So K cannot be reduced to −0.

Approach used in the below program is as follows

• We take an integer array ops[] initialized with random integers.

• Variable K is given a positive value.

• Function countOperations(int op[], int n, int k) takes K array Ops[] and its length as parameters and return operations required to reduce K to less than 0.

• Take the initial number of operations as 0 in count.

• Calculate sum of elements of ops[] and store in sum. If sum>=0 then return -1.

• If not while k>0 keep adding ops[i] and increment count. If k<0 break loop.

• Return count as result.

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
long countOperations(int op[], int n, int k){
   long count = 0;
   int sum=0;
   int i=0;
   for(int i=0;i<n;i++){
      sum+=op[i];
   }
   if(sum-k>=0)
      { return -1; } //number k can never be reduced as sum-k is always positive or 0
   while(k>0){
      for(i=0;i<n;i++){
         if(k>0){
            count++;
            k+=op[i];
         }
         else
            { break; }
      }
   }
   return count;
}
int main(){
   int Ops[] = { 1,-1,5,-11};
   int len= sizeof(Ops) / sizeof(Ops[0]);
   int K=10;
   long ans=countOperations(Ops,len,K);
   if(ans==-1)
      { cout<<"K cannot be reduced!!"; }
   else
      { cout<<"Number of operations : "<<ans; }
   return 0;
}

Output

If we run the above code it will generate the following output −

Number of operations : 8