416. Partition Equal Subset Sum¶

Difficulty: Medium

416. Partition Equal Subset Sum

Medium

Given an integer array nums, return true if you can partition the array into two subsets such that the sum of the elements in both subsets is equal or false otherwise.

Example 1:

Input: nums = [1,5,11,5]
Output: true
Explanation: The array can be partitioned as [1, 5, 5] and [11].

Example 2:

Input: nums = [1,2,3,5]
Output: false
Explanation: The array cannot be partitioned into equal sum subsets.

Constraints:

1 <= nums.length <= 200
1 <= nums[i] <= 100

Solution¶

class Solution {
    public boolean canPartition(int[] nums) {
        int sum = 0;
        int n = nums.length;
        for (int ele : nums) sum += ele;
        if (sum % 2 == 1) return false;
        else{
            int target = sum / 2;
            int dp[][] = new int[n + 1][target + 1];
            for(int temp[] : dp) Arrays.fill(temp , -1);
            if (solve(n - 1, nums, target, dp) == 1) return true;
            return false;
        }
    }
    public static int solve(int ind , int arr[],int target, int dp[][]){
        if(target == 0) return 1;
        if(ind == 0){
            if (arr[0] == target) return 1;
            return 0;
        }
        if(dp[ind][target] != -1)return dp[ind][target];
        int not_take = solve(ind - 1, arr ,target, dp);
        int take = 0;
        if(arr[ind] <= target){
            take = solve(ind - 1, arr, target - arr[ind],dp);
        }
        return dp[ind][target] = (take | not_take);
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here