Contains Duplicate II

Contains Duplicate II - Problem

Given an integer array nums and an integer k, return true if there are two distinct indices i and j in the array such that nums[i] == nums[j] and abs(i - j) <= k.

In other words, find if there are duplicate values within a window of size k.

Input & Output

Example 1 — Basic Case

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

› Output: true

💡 Note: There are two 1's at indices 0 and 3. The distance |3-0| = 3 which equals k=3, so return true.

Example 2 — Distance Too Large

$ Input: nums = [1,0,1,1], k = 1

› Output: true

💡 Note: There are two 1's at indices 2 and 3. The distance |3-2| = 1 which equals k=1, so return true.

Example 3 — No Duplicates Within k

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

› Output: false

💡 Note: No duplicate values exist within distance k=2. The closest duplicates are at distance 3.

Constraints

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

Visualization

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Asked in

a Amazon 12 Apple 8 M Microsoft 6 G Google 4

The key insight is to use a sliding window approach with a hash set to efficiently track duplicates within distance k. The optimal solution maintains a window of the most recent k elements, checking for duplicates as we slide the window. Time: O(n), Space: O(min(n,k))

Common Approaches

✓ Brute Force - Check All Pairs

⏱️ Time: O(n×k) Space: O(1)

For every element at index i, check all elements at indices from i+1 to min(i+k, n-1) to see if any match. This ensures we find duplicates within the required distance k.

Hash Map with Last Index

⏱️ Time: O(n) Space: O(n)

Use a hash map to store the most recent index where each element was seen. When we encounter an element, check if it was seen before and if the distance is within k. This approach remembers the last position of each element.

Sliding Window with Hash Set

⏱️ Time: O(n) Space: O(min(n,k))

Use a hash set to track elements within the current window of size k. Slide the window by adding new elements and removing old ones. If we try to add an element that already exists in the set, we found a duplicate within distance k.

Brute Force - Check All Pairs — Algorithm Steps

Iterate through each element
For each element, check the next k elements
Return true if any duplicate is found within distance k

Visualization

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Step-by-Step Walkthrough

Start

Begin with first element

Check

Compare with next k elements

Found

Return true if duplicate found

Code -

solution.c — C

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

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

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int k;
    scanf("%d", &k);
    
    bool result = solution(nums, numsSize, k);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×k)

For each of n elements, we check up to k subsequent elements

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for iteration

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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