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 { publ ic: 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 -
MY CODE VILLAGE
Code like a pro !