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  • Leetcode: Patching Array

    Given a sorted positive integer array nums and an integer n, add/patch elements to the array such that any number in range [1, n] inclusive can be formed by the sum of some elements in the array. Return the minimum number of patches required.
    
    Example 1:
    nums = [1, 3], n = 6
    Return 1.
    
    Combinations of nums are [1], [3], [1,3], which form possible sums of: 1, 3, 4.
    Now if we add/patch 2 to nums, the combinations are: [1], [2], [3], [1,3], [2,3], [1,2,3].
    Possible sums are 1, 2, 3, 4, 5, 6, which now covers the range [1, 6].
    So we only need 1 patch.
    
    Example 2:
    nums = [1, 5, 10], n = 20
    Return 2.
    The two patches can be [2, 4].
    
    Example 3:
    nums = [1, 2, 2], n = 5
    Return 0.

    reference: https://leetcode.com/discuss/82822/solution-explanation

    Let miss be the smallest sum in [1,n] that we might be missing. Meaning we already know we can build all sums in [1,miss). Then if we have a number num <= miss in the given array, we can add it to those smaller sums to build all sums in [1,miss+num). If we don't, then we must add such a number to the array, and it's best to add miss itself, to maximize the reach.

     1 public class Solution {
     2     public int minPatches(int[] nums, int n) {
     3         long missed=1;
     4         int added=0, i=0;
     5         while (missed <= n) {
     6             if (i<nums.length && nums[i]<=missed) {
     7                 missed = missed + nums[i];
     8             }
     9             else {
    10                 added++;
    11                 missed = missed + missed;
    12                 i--;
    13             }
    14             i++;
    15         }
    16         return added;
    17     }
    18 }
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/5165054.html
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