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  1. Prepare
  2. Algorithms
  3. Greedy
  4. Permuting Two Arrays
  5. Discussions

Permuting Two Arrays

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349 Discussions

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  • ahmet_alp77
     + 0 comments

    Java:

    public static String twoArrays(int k, List<Integer> A, List<Integer> B) {
    A.sort(Comparator.naturalOrder());
    B.sort(Comparator.reverseOrder());
    for (int i=0; i < A.size(); ++i) {
        if (k > A.get(i) + B.get(i)) {
            return "NO";
        }
    }
    return "YES";
}

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the vidéo explanation here : https://youtu.be/JQQV9IZlz7g

    string twoArrays(int k, vector<int> A, vector<int> B) {
    sort(A.begin(), A.end());
    sort(B.begin(), B.end(), [](int l, int r){ return l > r;});
    for(int i = 0; i < A.size(); i++){
        if(A[i] + B[i] < k) return "NO";
    }
    return "YES";
}

  • emhhacker
     + 0 comments

    My Java solution with o(n log n) time complexity and o(1) space complexity:

    public static String twoArrays(int k, List<Integer> A, List<Integer> B) {
        // goal: order arrs a and b to where a[i] + b[i] >= k
        
        //order arr a ascending, order arr b descending
        Collections.sort(A);
        Collections.sort(B, Collections.reverseOrder());
        
        //check each val
        for(int i = 0; i < A.size(); i++){
            if(A.get(i) + B.get(i) < k) return "NO"; 
        }
        return "YES"; //all vals of A and B were > k when summed together
    }

  • iremitaze
     + 0 comments

    C++ solution ==========>>>>>>>>>>>>>

    string twoArrays(int k, vector A, vector B) { if(A.size() != B.size()) { return "NO"; } int n = A.size(); sort(A.begin(), A.end()); sort(B.begin(), B.end(), greater());

    for(int i = 0; i < n; i++ ) {
    if(A[i] + B[i] < k) {
        return "NO";
    }
}
return "YES";

    }

  • patrickgao2020
     + 0 comments

    Here is my Python solution!

    def twoArrays(k, A, B):
    B.sort()
    for num1 in A:
        working = False
        for num2 in B:
            if num1 + num2 >= k:
                working = True
                B.remove(num2)
                break
        if not working:
            return "NO"
    return "YES"