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2485. Find the Pivot Integer | Binary search

 Given a positive integer n, find the pivot integer x such that:

  • The sum of all elements between 1 and x inclusively equals the sum of all elements between x and n inclusively.

Return the pivot integer x. If no such integer exists, return -1. It is guaranteed that there will be at most one pivot index for the given input.

 

Example 1:

Input: n = 8
Output: 6
Explanation: 6 is the pivot integer since: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 = 21.

Example 2:

Input: n = 1
Output: 1
Explanation: 1 is the pivot integer since: 1 = 1.

Example 3:

Input: n = 4
Output: -1
Explanation: It can be proved that no such integer exist.

 

Constraints:

  • 1 <= n <= 1000

Solution :

class Solution {
public:
    int pivotInteger(int n) {

        int sum = ((n)*(n+1))/2;
        int i = 1;
        int j = n ;
        while(i <= j )
        {
            int mid = (i +j)/2;
            int sum1 = (mid*(mid +1))/2;
            if(sum1 == sum - sum1 + mid)
            {
                return mid;
            }
            else
            if(sum1 < sum - sum1 + mid)
            {
                i = mid +1 ;
            }
            else
            {
                j = mid -1  ;
            }
        }
        return -1;
      
    }
};



Time complexity : O(log n )

Link :  https://leetcode.com/problems/find-the-pivot-integer
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