Contains Nearby Duplicate Window

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In Contains Nearby Duplicate Window, you are checking whether the same number appears again close enough in index distance. You get an integer array nums and an integer k, and the answer is a boolean. Return true only when there are two different indices i and j such that nums[i] == nums[j] and |i - j| <= k. If duplicates exist but the distance is larger than k, that still counts as false.

A practical way to solve this is to scan from left to right while storing the latest index for each value in a hash map. When a value appears, compare current index with its previous index immediately. If the gap is within k, you can stop early and return true. If not, update the stored index and continue.

The judge validates tight-window cases, repeated values that are too far apart, single-element arrays, and k = 0 behavior. Your output must always be a strict boolean result based on index distance, not just whether duplicates exist somewhere in the array.

Examples

Example 1

Input

nums = [1,2,3,1], k = 3

Output

true

Explanation

Value 1 appears at indices 0 and 3, distance 3.

Example 2

Input

nums = [1,0,1,1], k = 1

Output

true

Explanation

Last two ones are adjacent.

Example 3

Input

nums = [1,2,3,1,2,3], k = 2

Output

false

Explanation

No duplicate appears within distance 2.

Algorithm Flow

Recommendation Algorithm Flow for Contains Nearby Duplicate Window - Budibadu

Best Answers

java

import java.util.*;
class Solution {
    public boolean contains_nearby_duplicate_window(int[] nums, int k) {
        if (k <= 0) return false;
        Map<Integer, Integer> last = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            if (last.containsKey(nums[i]) && i - last.get(nums[i]) <= k) return true;
            last.put(nums[i], i);
        }
        return false;
    }
}

javascript

function contains_nearby_duplicate_window(nums, k) {
    if (k <= 0) return false;
    var last = {};
    for (var i = 0; i < nums.length; i++) {
        var key = String(nums[i]);
        if (last[key] !== undefined && i - last[key] <= k) return true;
        last[key] = i;
    }
    return false;
}

php

<?php
function contains_nearby_duplicate_window($nums, $k) {
    if ($k <= 0) return false;
    $last = [];
    $n = count($nums);
    for ($i = 0; $i < $n; $i++) {
        $key = strval($nums[$i]);
        if (array_key_exists($key, $last) && $i - $last[$key] <= $k) return true;
        $last[$key] = $i;
    }
    return false;
}
?>

python

from typing import List

def contains_nearby_duplicate_window(nums: List[int], k: int) -> bool:
    if k <= 0:
        return False
    last = {}
    for i, x in enumerate(nums):
        if x in last and i - last[x] <= k:
            return True
        last[x] = i
    return False

rust

use std::collections::HashMap;
fn contains_nearby_duplicate_window(nums: Vec<i32>, k: i32) -> bool {
    if k <= 0 { return false; }
    let mut last: HashMap<i32, i32> = HashMap::new();
    for (i, &x) in nums.iter().enumerate() {
        let idx = i as i32;
        if let Some(&prev) = last.get(&x) {
            if idx - prev <= k { return true; }
        }
        last.insert(x, idx);
    }
    false
}

typescript

function contains_nearby_duplicate_window(nums: number[], k: number): boolean {
    if (k <= 0) return false;
    var last: { [key: string]: number } = {};
    for (var i = 0; i < nums.length; i++) {
        var key = String(nums[i]);
        if (last[key] !== undefined && i - last[key] <= k) return true;
        last[key] = i;
    }
    return false;
}

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Contains Nearby Duplicate Window

Examples

Algorithm Flow

Best Answers

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