Find the Duplicate Number

Find the Duplicate Number - Problem

Given an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive.

There is only one repeated number in nums, return this repeated number.

You must solve the problem without modifying the array nums and using only constant extra space.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,3,4,3,2]

› Output: 3

💡 Note: The number 3 appears at indices 1 and 3, making it the duplicate number

Example 2 — Duplicate at End

$ Input: nums = [3,1,3,4,2]

› Output: 3

💡 Note: The number 3 appears at indices 0 and 2, so 3 is the duplicate

Example 3 — Minimum Size

$ Input: nums = [1,1]

› Output: 1

💡 Note: With only 2 elements in range [1,1], the duplicate must be 1

Constraints

1 ≤ n ≤ 10⁵
nums.length == n + 1
1 ≤ nums[i] ≤ n
All integers in nums appear only once except for precisely one integer which appears two or more times.

Visualization

Tap to expand

Asked in

M Microsoft 45 🍎 Apple 38 a Amazon 35 G Google 32

The key insight is to recognize this as a cycle detection problem by treating array values as pointers. The optimal approach uses Floyd's Cycle Detection Algorithm (Tortoise and Hare) to find the duplicate in O(n) time with O(1) space.

Common Approaches

✓ Binary Search on Answer Space

⏱️ Time: O(n log n) Space: O(1)

Since numbers are in range [1,n], we can binary search on this range. For each candidate value, count how many elements are ≤ that value. By pigeonhole principle, if count > candidate, the duplicate is in [1, candidate].

Brute Force - Check All Pairs

⏱️ Time: O(n²) Space: O(1)

Use nested loops to check all possible pairs of elements. When we find two elements at different indices with the same value, that's our duplicate.

Floyd's Cycle Detection (Tortoise and Hare)

⏱️ Time: O(n) Space: O(1)

Transform the problem into cycle detection in a linked list. Treat each array value as a pointer to the next index. The duplicate creates a cycle, and Floyd's algorithm finds the cycle start (which is our duplicate).

Binary Search on Answer Space — Algorithm Steps

Set search range [1, n]
For mid value, count elements ≤ mid
If count > mid, duplicate is in left half, else right half

Visualization

Tap to expand

Step-by-Step Walkthrough

Range

Search in range [1, n]

Count

Count elements ≤ mid

Narrow

Adjust search based on count

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int left = 1, right = numsSize - 1;
    
    while (left < right) {
        int mid = left + (right - left) / 2;
        int count = 0;
        for (int i = 0; i < numsSize; i++) {
            if (nums[i] <= mid) count++;
        }
        
        if (count > mid) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    
    return left;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int count = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n') {
            nums[count++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    printf("%d\n", solution(nums, count));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Binary search runs log n iterations, each iteration counts n elements

✓ Linear Growth

Space Complexity

O(1)

Only using constant variables for search bounds and counters

✓ Linear Space

78.0K Views

High Frequency

~25 min Avg. Time

1.8K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Binary Search on Answer Space — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler