Contains Duplicate

Contains Duplicate - Problem

Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.

This is a fundamental problem that tests your understanding of data structures and algorithms for detecting duplicates efficiently.

Input & Output

Example 1 — Contains Duplicate

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

› Output: true

💡 Note: The value 1 appears at both index 0 and index 3, so there is a duplicate.

Example 2 — All Distinct

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

› Output: false

💡 Note: All elements are distinct, no duplicates found.

Example 3 — Multiple Duplicates

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

› Output: true

💡 Note: Several values appear multiple times (1, 2, 3, and 4), so there are duplicates.

Constraints

1 ≤ nums.length ≤ 10⁵
-10⁹ ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

G Google 25 A Amazon 30 M Microsoft 20 🍎 Apple 15

The key insight is using a hash set to track seen elements in a single pass. The optimal approach is the hash set method with O(n) time and O(n) space complexity.

Common Approaches

✓ Brute Force - Compare All Pairs

⏱️ Time: O(n²) Space: O(1)

Use nested loops to compare each element with all other elements in the array. If any two elements are equal, return true. If no duplicates are found after checking all pairs, return false.

Hash Set - Optimal Solution

⏱️ Time: O(n) Space: O(n)

Iterate through the array once, adding each element to a hash set. Before adding, check if the element already exists in the set. If it does, we found a duplicate.

Sort First Approach

⏱️ Time: O(n log n) Space: O(1)

Sort the array so that duplicate elements become adjacent. Then iterate through the sorted array and check if any two consecutive elements are equal.

Brute Force - Compare All Pairs — Algorithm Steps

Iterate through each element with index i
For each element, compare it with all elements after it (index j > i)
If nums[i] equals nums[j], return true
If no duplicates found, return false

Visualization

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Step-by-Step Walkthrough

Setup

Start with first element, compare with all others

Compare

Check if any pair of elements are equal

Found

Return true when duplicate found

Code -

solution.c — C

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

bool solution(int* nums, int numsSize) {
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if (nums[i] == nums[j]) {
                return true;
            }
        }
    }
    return false;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int nums[1000];
    int size = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    bool result = solution(nums, size);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops compare each element with every other element

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Compare All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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