# C++ Program to count number of operations needed to place elements whose index is smaller than value

Suppose we have an array A with n elements. We can perform these operations any number of times −

• Select any positive integer k

• Select any position in the sequence and insert k into that position

• So, the sequence is changed, we proceed with this sequence in the next operation.

We have to find minimum number of operations needed to satisfy the condition: A[i] <= i for all i in range 0 to n-1.

So, if the input is like A = [1, 2, 5, 7, 4], then the output will be 3, because we can perform the operations like: [1,2,5,7,4] to [1,2,3,5,7,4] to [1,2,3,4,5,7,4] to [1,2,3,4,5,3,7,4].

## Steps

To solve this, we will follow these steps −

maxj := 0
n := size of A
for initialize i := 0, when i < n, update (increase i by 1), do:
maxj := maximum of maxj and (A[i] - i - 1)
return maxj

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;

int solve(vector<int> A) {
int maxj = 0;
int n = A.size();
for (int i = 0; i < n; i++) {
maxj = max(maxj, A[i] - i - 1);
}
return maxj;
}
int main() {
vector<int> A = { 1, 2, 5, 7, 4 };
cout << solve(A) << endl;
}

## Input

{ 1, 2, 5, 7, 4 }

## Output

3

Updated on: 04-Mar-2022

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