Recently one of my friends shared this problem with me. I really found this beautiful and thought of giving it a share.I won't spoil the essence of this problem, but definitely expecting some nice approaches and solutions in the comment.
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Recently one of my friends shared this problem with me. I really found this beautiful and thought of giving it a share.I won't spoil the essence of this problem, but definitely expecting some nice approaches and solutions in the comment.
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Auto comment: topic has been updated by cryo_coder123 (previous revision, new revision, compare).
dp[i] = the longest subsequence from the first i element which has the condition (xor = k) and the last element in the subsequence is arr[i]. then dp[i] = dp[last (arr[i] ^ k)] + 1, and you can save what is the last x.
First if k=0 we can see the answer is the frequency of the element which occurs maximum number of times in the array. Else we can do the following:
First keep track of vector of positions of each element in the array. It is easy to see that the required subsequence must be of the form [x, x^k, x, x^k.....] for some 1<=x<=10^6. So just brute force for each possible value of x and greedily check the subsequence with maximum length we can obtain by using the vector of positions for x and x^k.
Time complexity: O(10^6) + O(n)
Ya did the same thing... have a look at my piece of code