Partition Problem

Given a set of positive integers, find if the set can be partitioned into two subsets such that the sum of these two subsets are equal. 

DFS approach. Sort the array in an increasing order. Create two buckets to save the sum of the two partition. Every dfs call chooses a number in the set and add it to the corresponding buckets such that the sum of the bucket will not exceed sum/2. After that move index forward. 

public static boolean makesquare(int[] nums) {if (nums == null || nums.length < 2) return false;    int sum = 0;    for (int num : nums) sum += num;    if (sum % 2 != 0) return false;        Arrays.sort(nums);    return dfs(nums, new int[2], nums.length-1, sum / 2);}private static boolean dfs(int[] nums, int[] sums, int index, int target) {// all numbers are used if (index == -1) {// check if one of the sums equals to target    if (sums[0] == target)     return true;    return false;}for (int i = 0; i < 2; i++) {    // try to add current number to sums    if (sums[i] + nums[index] > target) continue;    sums[i] += nums[index];    // each number can only be used once, so we need to decrease index by 1 each time        if (dfs(nums, sums, index - 1, target)) return true;            // return values     sums[i] -= nums[index];}return false;}

DP approach(wiki). 

Let p(i, j) to be true if the sum(x0, x1, ... , xj) = i. 

Then the sub-problem would be,

the sum of the subset (x0, x1, .. , xj-1) equals to i, p(i, j-1) is true; 

or the sum of the subset (x0, x1, ... , xj-1) equals to i-xj, p(i-xj, j-1) is true.