package com.fishercoder.solutions;

import java.util.ArrayList;
import java.util.List;

/**
 * 330. 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.
 */
public class _330 {

    public static class Solution1 {
        /**
         * credit: https://leetcode.com/articles/patching-array/ and https://discuss.leetcode.com/topic/35494/solution-explanation/2
         *
         * Let miss be the smallest sum in [0,n] that we might be missing. Meaning we already know we
         * can build all sums in [0,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 [0,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.
         *
         * Example: Let's say the input is nums = [1, 2, 4, 13, 43] and n = 100. We need to ensure that
         * all sums in the range [1,100] are possible. Using the given numbers 1, 2 and 4, we can
         * already build all sums from 0 to 7, i.e., the range [0,8). But we can't build the sum 8, and
         * the next given number (13) is too large. So we insert 8 into the array. Then we can build all
         * sums in [0,16). Do we need to insert 16 into the array? No! We can already build the sum 3,
         * and adding the given 13 gives us sum 16. We can also add the 13 to the other sums, extending
         * our range to [0,29). And so on. The given 43 is too large to help with sum 29, so we must
         * insert 29 into our array. This extends our range to [0,58). But then the 43 becomes useful
         * and expands our range to [0,101). At which point we're done.
         */

        public int minPatches(int[] nums, int n) {
            long misses = 1;//use long to avoid integer addition overflow
            int patches = 0;
            int i = 0;
            while (misses <= n) {
                if (i < nums.length && nums[i] <= misses) { //miss is covered
                    misses += nums[i++];
                } else { //patch miss to the array
                    misses += misses;
                    patches++;//increase the answer
                }
            }
            return patches;
        }

        public List<Integer> findPatches(int[] nums, int n) {
            long misses = 1;//use long to avoid integer addition overflow
            List<Integer> patches = new ArrayList<>();
            int i = 0;
            while (misses <= n) {
                if (i < nums.length && nums[i] <= misses) { //miss is covered
                    misses += nums[i++];
                } else { //patch miss to the array
                    patches.add((int) misses);//increase the answer
                    misses += misses;
                }
            }
            return patches;
        }
    }

}

@@ -85,744 +85,8 @@
 t;%0A%0A
-/**%0A * 330. Patching Array%0A *%0A * Given a sorted positive integer array nums and an integer n,%0A * add/patch elements to the array such that any number in range %5B1, n%5D%0A * inclusive can be formed by the sum of some elements in the array. Return the minimum number of patches required.%0A%0A Example 1:%0A nums = %5B1, 3%5D, n = 6%0A Return 1.%0A%0A Combinations of nums are %5B1%5D, %5B3%5D, %5B1,3%5D, which form possible sums of: 1, 3, 4.%0A Now if we add/patch 2 to nums, the combinations are: %5B1%5D, %5B2%5D, %5B3%5D, %5B1,3%5D, %5B2,3%5D, %5B1,2,3%5D.%0A Possible sums are 1, 2, 3, 4, 5, 6, which now covers the range %5B1, 6%5D.%0A So we only need 1 patch.%0A%0A Example 2:%0A nums = %5B1, 5, 10%5D, n = 20%0A Return 2.%0A The two patches can be %5B2, 4%5D.%0A%0A Example 3:%0A nums = %5B1, 2, 2%5D, n = 5%0A Return 0.%0A */%0A
 publ
@@ -285,32 +285,36 @@
 ion/2%0A         *
+ %3Cp%3E
 %0A         * Let 
@@ -708,16 +708,16 @@
  reach.%0A
-
         
@@ -718,16 +718,20 @@
        *
+ %3Cp%3E
 %0A       

refactor 330