116 lines
4.3 KiB
C++
116 lines
4.3 KiB
C++
// Source : https://leetcode.com/problems/patching-array/
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// Author : Hao Chen
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// Date : 2016-03-01
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/***************************************************************************************
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*
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* Given a sorted positive integer array nums and an integer n, add/patch elements to
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* the array such that any number in range [1, n] inclusive can be formed by the sum of
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* some elements in the array. Return the minimum number of patches required.
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*
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* Example 1:
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* nums = [1, 3], n = 6
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* Return 1.
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*
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* Combinations of nums are [1], [3], [1,3], which form possible sums of: 1, 3, 4.
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* Now if we add/patch 2 to nums, the combinations are: [1], [2], [3], [1,3], [2,3],
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* [1,2,3].
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* Possible sums are 1, 2, 3, 4, 5, 6, which now covers the range [1, 6].
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* So we only need 1 patch.
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*
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* Example 2:
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* nums = [1, 5, 10], n = 20
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* Return 2.
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* The two patches can be [2, 4].
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*
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* Example 3:
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* nums = [1, 2, 2], n = 5
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* Return 0.
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*
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* Credits:Special thanks to @dietpepsi for adding this problem and creating all test
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* cases.
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***************************************************************************************/
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class Solution {
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public:
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int minPatches(vector<int>& nums, int n) {
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return minPatches_02(nums, n);
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return minPatches_01(nums, n);
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}
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// Greedy Algorithm
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// (Assume the array is sorted already)
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//
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// Let do some observation at first,
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//
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// 1) if we have [1,2] then we can cover 1, 2, 3
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// 2) if we have [1,2,3] then we can cover 1,2,3,4,5,6
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// So, it looks we can simply add all of nums together, then we can find out max number we can reach.
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//
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// 3) if we have [1,2,5], we can see
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// 3.1) [1,2] can cover 1,2,3, but we cannot reach 4,
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// 3.2) then we patch 4, then we have [1,2,4] which can cover 1,2,3(1+2),4,5(1+4),6(2+4), 7(1+2+4)
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// 3.3) we can see [1,2,4] can reach to 7 - sum all of them
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// 3.4) then [1,2,4,5], we can reach to 12 - 1,2,3,4,5,6,7,8(1+2+5),9(4+5),10(1+4+5), 11(2+4+5), 12(1+2+4+5)
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//
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// So, we can have figure out our solution
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//
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// 0) considering the `n` we need to cover.
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// 1) maintain a variable we are trying to patch, suppose named `try_patch`
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// 2) if `try_patch` >= nums[i] then, we just keep add the current array item,
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// and set the `try_patch` to the next patch candidate number - `sum+1`
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// 3) if `try_patch` < nums[i], which means we need to patch.
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//
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int minPatches_01(vector<int>& nums, int n) {
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long covered = 0; //avoid integer overflow
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int patch_cnt = 0;
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int i = 0;
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while (i<nums.size() ){
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// set the `try_patch` is the next number which we cannot cover
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int try_patch = covered + 1;
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// if the `try_patch` can cover the current item, then just sum it,
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// then we can have the max number we can cover so far
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if ( try_patch >= nums[i]) {
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covered += nums[i];
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i++;
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} else { // if the `try_patch` cannot cover the current item, then we find the number we need to patch
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patch_cnt++;
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//cout << "patch " << try_patch << endl;
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covered = covered + try_patch;
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}
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if (covered >=n) break;
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}
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//for the case - [1], 7
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//the above while-loop just process all of the numbers in the array,
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//but we might not reach the goal, so, we need keep patching.
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while (covered < n) {
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int try_patch = covered + 1;
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patch_cnt++;
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//cout << "patch " << try_patch << endl;
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covered = covered + try_patch;
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}
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return patch_cnt;
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}
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//The following solution just re-organizes the solution above, and make it shorter
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int minPatches_02(vector<int>& nums, int n) {
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long covered = 0;
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int patch_cnt = 0;
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int i = 0;
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while ( covered < n){
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if (i<nums.size() && nums[i] <= covered + 1) {
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covered += nums[i++];
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}else{
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//cout << "patch " << covered + 1 << endl;
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covered = 2 * covered + 1;
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patch_cnt++;
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}
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}
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return patch_cnt;
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}
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};
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