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1064. Fixed Point

Description

Given an array of distinct integers arr, where arr is sorted in ascending order, return the smallest index i that satisfies arr[i] == i. If there is no such index, return -1.

 

Example 1:

Input: arr = [-10,-5,0,3,7]
Output: 3
Explanation: For the given array, arr[0] = -10, arr[1] = -5, arr[2] = 0, arr[3] = 3, thus the output is 3.

Example 2:

Input: arr = [0,2,5,8,17]
Output: 0
Explanation: arr[0] = 0, thus the output is 0.

Example 3:

Input: arr = [-10,-5,3,4,7,9]
Output: -1
Explanation: There is no such i that arr[i] == i, thus the output is -1.

 

Constraints:

• 1 <= arr.length < 104
• -109 <= arr[i] <= 109

 

Follow up: The O(n) solution is very straightforward. Can we do better?

Solutions

Binary search.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public int fixedPoint(int[] arr) {
          int left = 0, right = arr.length - 1;
          while (left < right) {
              int mid = (left + right) >> 1;
              if (arr[mid] >= mid) {
                  right = mid;
              } else {
                  left = mid + 1;
              }
          }
          return arr[left] == left ? left : -1;
      }
  }
  

• class Solution {
  public:
      int fixedPoint(vector<int>& arr) {
          int left = 0, right = arr.size() - 1;
          while (left < right) {
              int mid = left + right >> 1;
              if (arr[mid] >= mid) {
                  right = mid;
              } else {
                  left = mid + 1;
              }
          }
          return arr[left] == left ? left : -1;
      }
  };
  

• class Solution:
      def fixedPoint(self, arr: List[int]) -> int:
          left, right = 0, len(arr) - 1
          while left < right:
              mid = (left + right) >> 1
              if arr[mid] >= mid:
                  right = mid
              else:
                  left = mid + 1
          return left if arr[left] == left else -1
  
  

• func fixedPoint(arr []int) int {
  	left, right := 0, len(arr)-1
  	for left < right {
  		mid := (left + right) >> 1
  		if arr[mid] >= mid {
  			right = mid
  		} else {
  			left = mid + 1
  		}
  	}
  	if arr[left] == left {
  		return left
  	}
  	return -1
  }
  

• function fixedPoint(arr: number[]): number {
      let left = 0;
      let right = arr.length - 1;
      while (left < right) {
          const mid = (left + right) >> 1;
          if (arr[mid] >= mid) {
              right = mid;
          } else {
              left = mid + 1;
          }
      }
      return arr[left] === left ? left : -1;
  }
  
  

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