Fixed Point

Fixed Point - Problem

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.

A fixed point in an array is an index where the value equals the index itself.

Input & Output

Example 1 — No Fixed Point

$ Input: arr = [-1,0,3,5,9,12]

› Output: -1

💡 Note: Check each index: arr[0]=-1≠0, arr[1]=0≠1, arr[2]=3≠2, arr[3]=5≠3, arr[4]=9≠4, arr[5]=12≠5. No fixed point exists.

Example 2 — Fixed Point Found

$ Input: arr = [0,2,3,4,5]

› Output: 0

💡 Note: arr[0] = 0, which equals the index 0. This is the smallest (and only) fixed point.

Example 3 — Multiple Fixed Points

$ Input: arr = [-1,1,3,4]

› Output: 1

💡 Note: arr[1] = 1, which equals the index 1. This is the only fixed point.

Constraints

1 ≤ arr.length ≤ 10⁴
-10⁹ ≤ arr[i] ≤ 10⁹
arr is sorted in ascending order
All elements in arr are distinct

Visualization

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Asked in

G Google 25 f Facebook 18 a Amazon 15 M Microsoft 12

The key insight is that since the array is sorted and contains distinct elements, we can use binary search to efficiently find the fixed point. The optimal approach uses binary search to achieve O(log n) time complexity by eliminating half the search space in each iteration. Time: O(log n), Space: O(1)

Common Approaches

✓ Binary Search

⏱️ Time: O(log n) Space: O(1)

Since the array is sorted, we can use binary search. If arr[mid] > mid, then all elements to the right will also be greater than their indices. If arr[mid] < mid, we need to search the right half.

Linear Search

⏱️ Time: O(n) Space: O(1)

Iterate through the array from left to right, checking if the value at each index equals the index itself. Return the first match found, or -1 if no match exists.

Binary Search — Algorithm Steps

Set left = 0, right = n-1
While left <= right, calculate mid
If arr[mid] == mid, we found a fixed point
If arr[mid] > mid, search left half (right = mid - 1)
If arr[mid] < mid, search right half (left = mid + 1)

Visualization

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Step-by-Step Walkthrough

Initial

Set left=0, right=n-1

Compare

Check arr[mid] vs mid

Eliminate

Remove half based on comparison

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* arr, int arrSize) {
    int left = 0, right = arrSize - 1;
    
    while (left <= right) {
        int mid = left + (right - left) / 2;
        
        if (arr[mid] == mid) {
            return mid;
        } else if (arr[mid] > mid) {
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    
    return -1;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array manually
    int arr[1000];
    int size = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n' && token[0] != ' ') {
            arr[size++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(arr, size);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

Binary search eliminates half the search space in each iteration

⚡ Linearithmic

Space Complexity

O(1)

Only using a constant amount of extra variables for pointers

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Binary Search — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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