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| 1 | +// Source : https://leetcode.com/problems/ways-to-split-array-into-three-subarrays/ |
| 2 | +// Author : Hao Chen |
| 3 | +// Date : 2021-05-11 |
| 4 | + |
| 5 | +/***************************************************************************************************** |
| 6 | + * |
| 7 | + * A split of an integer array is good if: |
| 8 | + * |
| 9 | + * The array is split into three non-empty contiguous subarrays - named left, mid, right |
| 10 | + * respectively from left to right. |
| 11 | + * The sum of the elements in left is less than or equal to the sum of the elements in mid, |
| 12 | + * and the sum of the elements in mid is less than or equal to the sum of the elements in right. |
| 13 | + * |
| 14 | + * Given nums, an array of non-negative integers, return the number of good ways to split nums. As the |
| 15 | + * number may be too large, return it modulo 10^9 + 7. |
| 16 | + * |
| 17 | + * Example 1: |
| 18 | + * |
| 19 | + * Input: nums = [1,1,1] |
| 20 | + * Output: 1 |
| 21 | + * Explanation: The only good way to split nums is [1] [1] [1]. |
| 22 | + * |
| 23 | + * Example 2: |
| 24 | + * |
| 25 | + * Input: nums = [1,2,2,2,5,0] |
| 26 | + * Output: 3 |
| 27 | + * Explanation: There are three good ways of splitting nums: |
| 28 | + * [1] [2] [2,2,5,0] |
| 29 | + * [1] [2,2] [2,5,0] |
| 30 | + * [1,2] [2,2] [5,0] |
| 31 | + * |
| 32 | + * Example 3: |
| 33 | + * |
| 34 | + * Input: nums = [3,2,1] |
| 35 | + * Output: 0 |
| 36 | + * Explanation: There is no good way to split nums. |
| 37 | + * |
| 38 | + * Constraints: |
| 39 | + * |
| 40 | + * 3 <= nums.length <= 10^5 |
| 41 | + * 0 <= nums[i] <= 10^4 |
| 42 | + ******************************************************************************************************/ |
| 43 | + |
| 44 | +constint MOD = (int) (1e9 +7); |
| 45 | + |
| 46 | +classSolution { |
| 47 | +public: |
| 48 | +intwaysToSplit(vector<int>& nums) { |
| 49 | +int len = nums.size(); |
| 50 | + vector<int>presum(len,0); |
| 51 | + presum[0] = nums[0]; |
| 52 | +for(int i=1; i<nums.size(); i++){ |
| 53 | + presum[i] = presum[i-1] + nums[i]; |
| 54 | + } |
| 55 | + |
| 56 | +returnwaysToSplit_BS(presum);// Binary Search |
| 57 | +//return waysToSplit_TLE(presum); // Time Limit Error |
| 58 | + } |
| 59 | + |
| 60 | +intbinary_search(vector<int>& presum,int left,int i,bool searchLeft ) { |
| 61 | +int len = presum.size(); |
| 62 | +int l = i, r = len-1; |
| 63 | +int res = -1; |
| 64 | +while(l <= r) { |
| 65 | +int m = l + (r - l) /2; |
| 66 | +// if search Left, let middle item belong to left |
| 67 | +// if search Right, let middle item belong to right |
| 68 | +int x = searchLeft?0 :1; |
| 69 | +int right = presum[len-1] - presum[m-x]; |
| 70 | +int mid = presum[m-x] - presum[i-1]; |
| 71 | + |
| 72 | +if (left <= mid && mid <= right) { |
| 73 | + res = m; |
| 74 | +if (searchLeft) r = m -1; |
| 75 | +else l = m +1; |
| 76 | + }elseif (left > mid) { |
| 77 | + l = m +1; |
| 78 | + }else { |
| 79 | + r = m -1; |
| 80 | + } |
| 81 | + |
| 82 | + } |
| 83 | +return res; |
| 84 | + } |
| 85 | +intwaysToSplit_BS(vector<int>& presum) { |
| 86 | +int len = presum.size(); |
| 87 | +long cnt =0; |
| 88 | +for(int i=0; i<len-2; i++){ |
| 89 | +if (presum[i] > (presum[len-1] - presum[i]) /2)break; |
| 90 | +//find the most right position |
| 91 | +long l =binary_search(presum, presum[i], i+1,true); |
| 92 | +//find the most right position |
| 93 | +long r =binary_search(presum, presum[i], i+1,false); |
| 94 | +if (l == -1 || r == -1 )continue; |
| 95 | + cnt += (r-l); |
| 96 | +//cout << i << " - [" << l << "," << r << "]" << endl; |
| 97 | + } |
| 98 | +//cout << endl; |
| 99 | +return cnt % MOD; |
| 100 | + } |
| 101 | + |
| 102 | +intwaysToSplit_TLE(vector<int>& presum) { |
| 103 | +int len = presum.size(); |
| 104 | +int cnt =0; |
| 105 | +int left, mid, right; |
| 106 | +for(int i=0; i<len-2; i++){ |
| 107 | + left = presum[i]; |
| 108 | +for (int j=i+1; j<len-1; j++) { |
| 109 | + mid = presum[j] - presum[i]; |
| 110 | + right = presum[len-1] - presum[j]; |
| 111 | +if (left <= mid && mid <= right) { |
| 112 | + cnt++; |
| 113 | + } |
| 114 | + } |
| 115 | + } |
| 116 | +return cnt; |
| 117 | + } |
| 118 | +}; |