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DFS + memo. 阅读全文
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Actually, it is a Greedy problem : ) 阅读全文
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It is marked as Recursion on HR, but the optimal solution is DP. Step by step, discover internal mechanism. 阅读全文
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The trick of this problem is.. Adding 1+ coins to one pile, means nothing - you add one, then i just remove it.. so it is back to original game. So, b 阅读全文
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A simple NIM game in disguise: move 1 coin to one of previous slot, equals to removing.. Then no surprise, Sprague-Grundy theorem solves the problem. 阅读全文
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Another easy one solved by Sprague-Grundy theorem. Each pile is a sub-game, so you need to XOR SG value of all substates - since you can move # of 1 - 阅读全文
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A matter of OBSERVATION.. draw a triangle of parity, and watch. 阅读全文
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Discrete thinking - Play & Observe! " you can move numbers in any quadrant at the following positions to [x][y]: [x][y] or // upper left quadrant [x][ 阅读全文
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Another intuitive of application of Sprague-Grundy theorem. 阅读全文
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An intuitive problem for learning & applying Sprague-Grudy theorem: https://zhuanlan.zhihu.com/maigo/20611132 Lesson learnt, in Sprague-Grundy theorem 阅读全文
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The major trick is from another similar HR problem: subarray with max XOR: Build a binary tree bit by bit, and go from MSB to LSB, greedily. In this o 阅读全文
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Fun Greedy. My first thought was a DFS based solution... however the editorial provides a super neat 2-pass O(n) solution: Pass 1: if s[l] != s[r], ch 阅读全文
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Over-thinking will not get you there though : ) The trick is QUITE easy: # of unique chars. 阅读全文
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Nice problem that can help you understand 2 key skills below: - How to choose between Greedy and DP Is the local optimal the ONLY option? Can other ch 阅读全文
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Swapping - is actually a useless mask to the sln of this problem. The invariance of this problem is, swapping doesn't change # of balls in each contai 阅读全文
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Hey I found problems on HackerRank is more interesting than ones on LeetCode.. Strategy: observe\analyze bit by bit - DISCRETE THINKING. The idea is l 阅读全文
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A different approach to the problem, besides the popular math solution. 阅读全文
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A variation of 'maximum sum of subarray' - using hashmap 阅读全文
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Play it in your mind.. or it may be hard to reach to this intuitive solution... 阅读全文
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My first try was very close to a final solution .. however, this is a much neater solution: https://discuss.leetcode.com/topic/69436/concise-c-o-1-sol 阅读全文