[LeetCode] 416. Partition Equal Subset Sum

Problem

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.

Solution

DFS1

class Solution {
    public boolean canPartition(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();
        int sum = 0;
        for (int num: nums) {
            map.put(num, map.getOrDefault(num, 0)+1);
            sum += num;
        }
        if (sum % 2 != 0) return false;
        return dfs(map, sum/2);
    }
    private boolean dfs(Map<Integer, Integer> map, int target) {
        if (map.containsKey(target) && map.get(target) > 0) return true;
        for (int num: map.keySet()) {
            if (num < target && map.get(num) > 0) {
                map.put(num, map.get(num)-1);
                if (dfs(map, target-num)) return true;
                map.put(num, map.get(num)+1);
            }
        }
        return false;
    }
}

DFS2 - TLE

class Solution {
    public boolean canPartition(int[] nums) {
        int sum = 0;
        for (int num: nums) sum += num;
        if (sum%2 != 0) return false;
        int target = sum/2;
        return dfs(nums, new boolean[nums.length], 0, 0, target);
    }
    private boolean dfs(int[] nums, boolean[] used, int start, int sum, int target) {
        if (sum > target || start >= nums.length) return false;
        if (sum == target) return true;
        for (int i = start; i < nums.length; i++) {
            if (nums[i] > target) return false;
            if (!used[i]) {
                used[i] = true;
                if (dfs(nums, used, i+1, sum+nums[i], target)) return true;
                used[i] = false;
            }
        }
        return false;
    }
}

TLE for this test case:
[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,100]