Bit Manipulation - Medium Level

 Leetcode 416 Partition Equal Subset Sum

Given a non-empty array nums 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.

Constraints:

1 <= nums.length <= 200

1 <= nums[i] <= 100

Analysis:

There are different ways to solve this problem, such as dp with a time complexity of O(N^2).

Since N is small to this question, so it is Okay to pass OJ.

Besides the small N, the value of each element is also very small. So this gives us some chance to use "space to trad off time".

The data structure to be used is bitset. For bitset, each bit can be either 0 or 1. The index of that bit can be used as the corresponding sum. When the bit is 1, means there is a sum with the value of its index.

When a new number comes, this number needs to be added to all the previous sums, to form new "previous" sums. Thus for each number, we need to go through all the previous sums, the time complexity of which is O(N^2) in average. There are N numbers, so the overall time complexity is O(N^3).

With a bitset, for each number, we just need to use the shift operator (<< and >>) and the or (|) operator directly, without go through all the previous sums. The time complexity is from O(N^2) to O(1). Here is how it works:

1. All the previous sums are stored in the bitset (the index of 1's);

2. When a new number is taken into account, we just need to shift the bitset to the left by the value of the number (let say X). Then all the 1's will be shifted left by X bits, so all the previous sums are updated by adding X with this one time operation!

3. We still need to keep the previous sums before the update in 2. And or (|) operation between the bitset before the update and that after the update would perfectly solve this problem.

After scanning to the last number, we just need to check

1. the overall sum is even or not; if odd, return false directly

2. if even, we just need to check the bit value of the bit at the index of (total sum)/2 in the bitset. If 1, return true; o/w, return false.

See the code below:

class Solution {
public:
    bool canPartition(vector<int>& nums) {
        int sum = 0;
        bitset<20001> bt(1); // bt[0] is 1
        for(auto &a : nums) {
            sum += a;
            bt |= bt << a;
        }
        return !(sum & 1) && bt[sum >> 1];
    }
};

Upper Layer