416. Partition Equal Subset Sum
https://leetcode.com/problems/partition-equal-subset-sum/
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Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.
Note:
- Each of the array element will not exceed 100.
- The array size will not exceed 200.
Example 1:
Input: [1, 5, 11, 5] Output: true Explanation: The array can be partitioned as [1, 5, 5] and [11].
Example 2:
Input: [1, 2, 3, 5] Output: false Explanation: The array cannot be partitioned into equal sum subsets.
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| class Solution { | |
| public boolean canPartition(int[] nums) { | |
| int sum = 0; | |
| for (int num: nums) { | |
| sum += num; | |
| } | |
| if (sum % 2 == 1) { | |
| return false; | |
| } | |
| Boolean[][] memo = new Boolean[nums.length][1 + sum / 2]; | |
| return dfs(nums, 0, 0, sum / 2, memo); | |
| } | |
| private boolean dfs(int[] nums, int i, int prefix, int target, Boolean[][] memo) { | |
| if (i == nums.length) { | |
| return prefix == target; | |
| } | |
| if (memo[i][prefix] != null) { | |
| return memo[i][prefix]; | |
| } | |
| boolean ans = false; | |
| // include current element, only if inclusion is valid | |
| if (prefix + nums[i] <= target) { | |
| ans = dfs(nums, i + 1, prefix + nums[i], target, memo); | |
| } | |
| // skip current element | |
| ans = ans || dfs(nums, i + 1, prefix, target, memo); | |
| memo[i][prefix] = ans; | |
| return ans; | |
| } | |
| } |
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| class Solution { | |
| public boolean canPartition(int[] nums) { | |
| int sum = 0; | |
| for (int num: nums) { | |
| sum += num; | |
| } | |
| if (sum % 2 == 1) { | |
| return false; | |
| } | |
| return dfs(nums, 0, 0, sum / 2); | |
| } | |
| private boolean dfs(int[] nums, int i, int prefix, int target) { | |
| if (i == nums.length) { | |
| return prefix == target; | |
| } | |
| boolean ans = false; | |
| // include current element, only if inclusion is valid | |
| if (prefix + nums[i] <= target) { | |
| ans = dfs(nums, i + 1, prefix + nums[i], target); | |
| } | |
| // skip current element | |
| ans = ans || dfs(nums, i + 1, prefix, target); | |
| return ans; | |
| } | |
| } |
