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Pairs

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  • anarod_val25
     + 0 comments
    int pairs(int k, vector<int> arr) {
    sort(arr.begin(),arr.end());
    set<int> arrset;
    int pairs = 0;
    
    for(int n : arr){
        if(arrset.count(n+k)||arrset.count(n-k)){
            pairs++;
        }
        arrset.insert(n);
    }
    
    return pairs;
}

  • maciejka_67
     + 0 comments
    def pairs(k, arr):
    s1 = set(arr)
    s2 = set(i + k for i in s1)
    return len(s1 & s2)

  • h24020066
     + 0 comments

    java public static int pairs(int k, List arr) { int count = 0; int start = 0; Collections.sort(arr); for(int i = 1; i < arr.size(); i++){ int diff = arr.get(i) - arr.get(start); if(diff > k) { start++; i--; } else if (diff == k) { count++; start++; } } return count; }

  • jerrythea
     + 0 comments

    Java solution using Lambdas

    public static int pairs(int k, List<Integer> arr) {
    // Write your code here
    Set<Integer> complementSet = ConcurrentHashMap.newKeySet();
    arr.stream().forEach(f -> complementSet.add(f));
    return (int) arr.parallelStream().filter(d -> complementSet.contains(k + d)).count();
}

    }

  • 0oAimZo0
     + 0 comments

    Golang Time Complexity: O(n) Space Complexity: O(n)

    First: populates a hash map with all elements from the input array arr for quick lookups.

    Second: For each element x, it checks if x - k and x + k exist in the hash map. if exist, count

    The current element x is then removed from the hash map to prevent counting the same pair twice (e.g., counting (3, 5) when x=3 and then again when x=5)

    func pairs(k int32, arr []int32) int32 {
exist := make(map[int32]bool)
for _, x := range arr {
    exist[x] = true
}

count := int32(0)
for _, x := range arr {
    if exist[x-k] {
        count++
    }
    if exist[x+k] {
        count++
    }
    exist[x] = false
}
return count

    }

    `