416. 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:

1. Each of the array element will not exceed 100.

2. 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

Base case: dp[0][0] is true; (zero number consists of sum 0 is true)

Transition function:

For each number, if we don't pick it, dp[i][j] = dp[i-1][j], which means if the first i-1th elements has made it to j, dp[i][j] would also make it to j (we can just ignore nums[i]).

If we pick nums[i]. dp[i][j] = dp[i-1][j-nums[i]], which represents that j is composed of the current value nums[i] and the remaining composed of other previous numbers. Thus, the transition function is dp[i][j] = dp[i-1][j] || dp[i-1][j-nums[i]]

talking is cheap:

public boolean canPartition(int[] nums) {
    int sum = 0;
    
    for (int num : nums) {
        sum += num;
    }
    
    if ((sum & 1) == 1) {
        return false;
    }
    sum /= 2;

    int n = nums.length;
    boolean[][] dp = new boolean[n+1][sum+1];
    for (int i = 0; i < dp.length; i++) {
        Arrays.fill(dp[i], false);
    }
    
    dp[0][0] = true;
    
    for (int i = 1; i < n+1; i++) {
        dp[i][0] = true;
    }
    for (int j = 1; j < sum+1; j++) {
        dp[0][j] = false;
    }
    
    for (int i = 1; i < n+1; i++) {
        for (int j = 1; j < sum+1; j++) {
            dp[i][j] = dp[i-1][j];
            if (j >= nums[i-1]) {
                dp[i][j] = (dp[i][j] || dp[i-1][j-nums[i-1]]); // dp[i-1]表示不考慮nums[i]
            }
        }
    }
   
    return dp[n][sum];
}