LeetCode #287 — MEDIUM

Find the Duplicate Number

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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.

Example 1:

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

Example 2:

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

Example 3:

Input: nums = [3,3,3,3,3]
Output: 3

Constraints:

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

Follow up:

How can we prove that at least one duplicate number must exist in nums?
Can you solve the problem in linear runtime complexity?

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Binary Search · Bit Manipulation

Example 1

[1,3,4,2,2]

Example 2

[3,1,3,4,2]

Example 3

[3,3,3,3,3]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #287: Find the Duplicate Number
class Solution {
    public int findDuplicate(int[] nums) {
        int l = 0, r = nums.length - 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            int cnt = 0;
            for (int v : nums) {
                if (v <= mid) {
                    ++cnt;
                }
            }
            if (cnt > mid) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #287: Find the Duplicate Number
func findDuplicate(nums []int) int {
	return sort.Search(len(nums), func(x int) bool {
		cnt := 0
		for _, v := range nums {
			if v <= x {
				cnt++
			}
		}
		return cnt > x
	})
}

# Accepted solution for LeetCode #287: Find the Duplicate Number
class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        def f(x: int) -> bool:
            return sum(v <= x for v in nums) > x

        return bisect_left(range(len(nums)), True, key=f)

// Accepted solution for LeetCode #287: Find the Duplicate Number
impl Solution {
    #[allow(dead_code)]
    pub fn find_duplicate(nums: Vec<i32>) -> i32 {
        let mut left = 0;
        let mut right = nums.len() - 1;

        while left < right {
            let mid = (left + right) >> 1;
            let cnt = nums.iter().filter(|x| **x <= (mid as i32)).count();
            if cnt > mid {
                right = mid;
            } else {
                left = mid + 1;
            }
        }

        left as i32
    }
}

// Accepted solution for LeetCode #287: Find the Duplicate Number
function findDuplicate(nums: number[]): number {
    let l = 0;
    let r = nums.length - 1;
    while (l < r) {
        const mid = (l + r) >> 1;
        let cnt = 0;
        for (const v of nums) {
            if (v <= mid) {
                ++cnt;
            }
        }
        if (cnt > mid) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.