473. Matchsticks to Square
https://leetcode.com/problems/matchsticks-to-square/
Remember the story of Little Match Girl? By now, you know exactly what matchsticks the little match girl has, please find out a way you can make one square by using up all those matchsticks. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.
Your input will be several matchsticks the girl has, represented with their stick length. Your output will either be true or false, to represent whether you could make one square using all the matchsticks the little match girl has.
Example 1:
Input: [1,1,2,2,2] Output: true Explanation: You can form a square with length 2, one side of the square came two sticks with length 1.
Example 2:
Input: [3,3,3,3,4] Output: false Explanation: You cannot find a way to form a square with all the matchsticks.
Note:
- The length sum of the given matchsticks is in the range of
0to10^9. - The length of the given matchstick array will not exceed
15.
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Intuition
If the sum of all elements in array % 4 != 0
return false
DFS on all combinations, and keep track of sums
Terminate on last index .. reached here only if all sums are valid
return true
for (i = 0... 4)
if (sums[i] + nums[index] <= target)
sums[i] += nums[index]
if (dfs(nums, sums, index + 1, target))
return true
sums[i] -= nums[index]
return false
This tries all combinations
Can optimize
Sort the array descending first
It exits early if any sum > target
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Time - O(4 ^ N)
Space - O(N)
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Related problems
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| class Solution { | |
| public boolean makesquare(int[] nums) { | |
| if (nums == null || nums.length == 0) { | |
| return false; | |
| } | |
| int sum = 0; | |
| for (int num: nums) { | |
| sum += num; | |
| } | |
| if (sum % 4 != 0) { | |
| return false; | |
| } | |
| // Optimization so invalid sums i.e, sum > target => terminate earlier | |
| Arrays.sort(nums); | |
| reverse(nums); | |
| return dfs(nums, 0, new int[4], sum / 4); | |
| } | |
| private boolean dfs(int[] nums, int index, int[] sums, int target) { | |
| // Terminate, reached here only if sums were valid | |
| if (index == nums.length) { | |
| return true; | |
| } | |
| for (int i = 0; i < sums.length; i++) { | |
| if (sums[i] + nums[index] <= target) { | |
| sums[i] += nums[index]; | |
| if (dfs(nums, index + 1, sums, target)) { | |
| return true; | |
| } | |
| // backtrack to add nums[index] to other sums | |
| sums[i] -= nums[index]; | |
| } | |
| } | |
| return false; | |
| } | |
| private void reverse(int[] nums) { | |
| int i = 0, j = nums.length - 1; | |
| while (i < j) { | |
| int temp = nums[j]; | |
| nums[j] = nums[i]; | |
| nums[i] = temp; | |
| i++; | |
| j--; | |
| } | |
| } | |
| } |
