C++ program: Calculate the number of operations required to place an element with an index less than a value

Suppose we have an array A containing n elements. We can perform these operations multiple times -

• Select any positive integer k

• Select any position and insert k

  # at that position
• ##In this way, the sequence changes and we continue the sequence in the next operation.

ul>We must find the minimum number of operands required to satisfy the condition: A[i] So if the input is say A = [1, 2, 5, 7, 4] then the output will be 3 because we can do something 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

h2>To solve this problem we will follow the following 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 have a look at the following implementation for better Understand −

#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

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