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