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30 | 30 | */
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31 | 31 | public class _330 {
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32 | 32 |
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33 |
| - /**credit: https://leetcode.com/articles/patching-array/ and https://discuss.leetcode.com/topic/35494/solution-explanation/2 |
34 |
| - * |
35 |
| - * Let miss be the smallest sum in [0,n] that we might be missing. |
36 |
| - * Meaning we already know we can build all sums in [0,miss). |
37 |
| - * Then if we have a number num <= miss in the given array, |
38 |
| - * we can add it to those smaller sums to build all sums in [0,miss+num). |
39 |
| - * If we don't, then we must add such a number to the array, |
40 |
| - * and it's best to add miss itself, to maximize the reach. |
| 33 | + public static class Solution1 { |
| 34 | + /** |
| 35 | + * credit: https://leetcode.com/articles/patching-array/ and https://discuss.leetcode.com/topic/35494/solution-explanation/2 |
| 36 | + * |
| 37 | + * Let miss be the smallest sum in [0,n] that we might be missing. Meaning we already know we |
| 38 | + * can build all sums in [0,miss). Then if we have a number num <= miss in the given array, we |
| 39 | + * can add it to those smaller sums to build all sums in [0,miss+num). If we don't, then we must |
| 40 | + * add such a number to the array, and it's best to add miss itself, to maximize the reach. |
| 41 | + * |
| 42 | + * Example: Let's say the input is nums = [1, 2, 4, 13, 43] and n = 100. We need to ensure that |
| 43 | + * all sums in the range [1,100] are possible. Using the given numbers 1, 2 and 4, we can |
| 44 | + * already build all sums from 0 to 7, i.e., the range [0,8). But we can't build the sum 8, and |
| 45 | + * the next given number (13) is too large. So we insert 8 into the array. Then we can build all |
| 46 | + * sums in [0,16). Do we need to insert 16 into the array? No! We can already build the sum 3, |
| 47 | + * and adding the given 13 gives us sum 16. We can also add the 13 to the other sums, extending |
| 48 | + * our range to [0,29). And so on. The given 43 is too large to help with sum 29, so we must |
| 49 | + * insert 29 into our array. This extends our range to [0,58). But then the 43 becomes useful |
| 50 | + * and expands our range to [0,101). At which point we're done. |
| 51 | + */ |
41 | 52 |
|
42 |
| - Example: Let's say the input is nums = [1, 2, 4, 13, 43] and n = 100. |
43 |
| - We need to ensure that all sums in the range [1,100] are possible. |
44 |
| - Using the given numbers 1, 2 and 4, we can already build all sums from 0 to 7, i.e., the range [0,8). |
45 |
| - But we can't build the sum 8, and the next given number (13) is too large. |
46 |
| - So we insert 8 into the array. Then we can build all sums in [0,16). |
47 |
| - Do we need to insert 16 into the array? No! We can already build the sum 3, |
48 |
| - and adding the given 13 gives us sum 16. |
49 |
| - We can also add the 13 to the other sums, extending our range to [0,29). |
50 |
| - And so on. The given 43 is too large to help with sum 29, so we must insert 29 into our array. |
51 |
| - This extends our range to [0,58). |
52 |
| - But then the 43 becomes useful and expands our range to [0,101). |
53 |
| - At which point we're done.*/ |
54 |
| - |
55 |
| - public int minPatches(int[] nums, int n) { |
56 |
| - long misses = 1;//use long to avoid integer addition overflow |
57 |
| - int patches = 0; |
58 |
| - int i = 0; |
59 |
| - while (misses <= n) { |
60 |
| - if (i < nums.length && nums[i] <= misses) { //miss is covered |
61 |
| - misses += nums[i++]; |
62 |
| - } else { //patch miss to the array |
63 |
| - misses += misses; |
64 |
| - patches++;//increase the answer |
| 53 | + public int minPatches(int[] nums, int n) { |
| 54 | + long misses = 1;//use long to avoid integer addition overflow |
| 55 | + int patches = 0; |
| 56 | + int i = 0; |
| 57 | + while (misses <= n) { |
| 58 | + if (i < nums.length && nums[i] <= misses) { //miss is covered |
| 59 | + misses += nums[i++]; |
| 60 | + } else { //patch miss to the array |
| 61 | + misses += misses; |
| 62 | + patches++;//increase the answer |
| 63 | + } |
65 | 64 | }
|
| 65 | + return patches; |
66 | 66 | }
|
67 |
| - return patches; |
68 |
| - } |
69 | 67 |
|
70 |
| - public List<Integer> findPatches(int[] nums, int n) { |
71 |
| - long misses = 1;//use long to avoid integer addition overflow |
72 |
| - List<Integer> patches = new ArrayList<>(); |
73 |
| - int i = 0; |
74 |
| - while (misses <= n) { |
75 |
| - if (i < nums.length && nums[i] <= misses) { //miss is covered |
76 |
| - misses += nums[i++]; |
77 |
| - } else { //patch miss to the array |
78 |
| - patches.add((int) misses);//increase the answer |
79 |
| - misses += misses; |
| 68 | + public List<Integer> findPatches(int[] nums, int n) { |
| 69 | + long misses = 1;//use long to avoid integer addition overflow |
| 70 | + List<Integer> patches = new ArrayList<>(); |
| 71 | + int i = 0; |
| 72 | + while (misses <= n) { |
| 73 | + if (i < nums.length && nums[i] <= misses) { //miss is covered |
| 74 | + misses += nums[i++]; |
| 75 | + } else { //patch miss to the array |
| 76 | + patches.add((int) misses);//increase the answer |
| 77 | + misses += misses; |
| 78 | + } |
80 | 79 | }
|
| 80 | + return patches; |
81 | 81 | }
|
82 |
| - return patches; |
83 | 82 | }
|
84 | 83 |
|
85 | 84 | }
|
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