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  1. Prepare
  2. Data Structures
  3. Stacks
  4. AND xor OR
  5. Discussions

AND xor OR

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153 Discussions

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  • dehaka
     + 1 comment

    O(n), 3 pass over the array A. there are only linearly many pairs (M1, M2)

    int andXorOr(int N, const vector<int>& A) {
    int maxXOR = 0;
    vector<int> prevSmaller(N, -1), nextSmaller(N, N);
    stack<int> prev, next;
    prev.push(0);
    next.push(N - 1);
    for (int i = 1; i < N; i++) {
        while (!prev.empty() and A[prev.top()] >= A[i]) prev.pop();
        if (!prev.empty()) prevSmaller[i] = prev.top();
        prev.push(i);
    }
    for (int i = N - 2; i >= 0; i--) {
        while (!next.empty() and A[i] <= A[next.top()]) next.pop();
        if (!next.empty()) nextSmaller[i] = next.top();
        next.push(i);
    }
    for (int i = 0; i < N; i++) {
        if (prevSmaller[i] != -1) maxXOR = max(maxXOR, A[i] ^ A[prevSmaller[i]]);
        if (nextSmaller[i] != N) maxXOR = max(maxXOR, A[i] ^ A[nextSmaller[i]]);
    }
    return maxXOR;
}

    • pr3574
       + 0 comments

      andXorOr(a.size(),a)

      in main function where this function is being called, you have to pass 2 arguments one is length of that vector and the vector itself

  • zallesov
     + 1 comment

    How is it that 9 and 6 are the smallest and the next smallest numbers in the given array? I struggle to understand this part of the assignment ` Let M0 and M1 be the smallest and the next smallest element in the interval. What defines L and M in this case? Is it just the 0 and 1 values in the whole array?

    • pavankumarbugat1
       + 0 comments

      when we take array[0],array[1] ie., [9,6] we get maximum value for the given condition

  • gfostick
     + 1 comment
    def andXorOr(a):
    m = 0
    s = []
    for i in a:     
        while s and s[-1] >= i:
            m = max(m, i^s.pop())
        if s:
            m = max(m, i^s[-1])
        s.append(i)
    return m

    • jagadeeshwarans3
       + 0 comments

      great logic

  • omar2004
     + 0 comments

    My JS Solution

    function andXorOr(A) {
    var stack = [A[0], A[1]];
    var S = A[0] ^ A[1];
    
    for(let i = 2; i < A.length; i++){
        while(stack.length > 0 && stack.slice(-1) >= A[i]){
            S = Math.max(S, stack.slice(-1) ^ A[i]);
            stack.pop();
        }
        if(stack.length > 0) S = Math.max(S, stack.slice(-1) ^ A[i]);
        stack.push(A[i])
    }
    return S;
}

  • omar2004
     + 0 comments

    My Python Solution

    def andXorOr(a):
    stack = [a[0], a[1]]
    S = a[0] ^ a[1]
    for i in range(2, len(a)):
        while len(stack) > 0 and stack[-1] >= a[i]:
            S = max(S, stack[-1] ^ a[i])
            stack.pop()
        if len(stack) > 0:
            S = max(S, stack[-1] ^ a[i])
        stack.append(a[i])
    return S

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