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Formatted question description: https://leetcode.ca/all/2485.html

2485. Find the Pivot Integer

• Difficulty: Easy.
• Related Topics: Math, Prefix Sum.
• Similar Questions: Bulb Switcher.

Problem

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 (Java, C++, Python)

• Java
• C++
• Python
• Go
• TypeScript
• Php
• RenderScript

• class Solution {
      public int pivotInteger(int n) {
          for (int x = 1; x < 1000; ++x) {
              if ((1 + x) * x == (x + n) * (n - x + 1)) {
                  return x;
              }
          }
          return -1;
      }
  }
  

• class Solution {
  public:
      int pivotInteger(int n) {
          for (int x = 1; x < 1000; ++x) {
              if ((1 + x) * x == (x + n) * (n - x + 1)) {
                  return x;
              }
          }
          return -1;
      }
  };
  

• class Solution:
      def pivotInteger(self, n: int) -> int:
          for x in range(1, 1000):
              if (1 + x) * x == (x + n) * (n - x + 1):
                  return x
          return -1
  
  

• func pivotInteger(n int) int {
  	for x := 1; x < 1000; x++ {
  		if (1+x)*x == (x+n)*(n-x+1) {
  			return x
  		}
  	}
  	return -1
  }
  

• function pivotInteger(n: number): number {
      const y = Math.floor((n * (n + 1)) / 2);
      const x = Math.floor(Math.sqrt(y));
      return x * x === y ? x : -1;
  }
  
  

• class Solution {
      /**
       * @param Integer $n
       * @return Integer
       */
      function pivotInteger($n) {
          $sum = ($n * ($n + 1)) / 2;
          $pre = 0;
          for ($i = 1; $i <= $n; $i++) {
              if ($pre + $i === $sum - $pre) {
                  return $i;
              }
              $pre += $i;
          }
          return -1;
      }
  }
  
  

• impl Solution {
      pub fn pivot_integer(n: i32) -> i32 {
          let y = n * (n + 1) / 2;
          let x = (y as f64).sqrt() as i32;
  
          if x * x == y {
              return x;
          }
  
          -1
      }
  }
  

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).

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