Skip to main content

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

Comments

Post a Comment

Popular posts from this blog

Binary Search - Hard Level - Question 3

Binary Search - Hard Level - Question 3 878. Nth Magical Number A positive integer is magical if it is divisible by either a or b. Given the three integers n, a, and b, return the nth magical number. Since the answer may be very large, return it modulo 10^9 + 7. Analysis: Let us consider some examples first. Example 1, a = 4, b = 2. If b is dividable by a, then all the numbers which is dividable by a should be dividable by b as well. So the nth magical number should be n*b; Example 2, a = 3, b = 2. The multiples of 2 are: 2, 4, 6, 8, 10, 12, ... The multiple of 3 are: 3, 6, 9, 12, ... So the overlap is related to the minimum common multiple between a and b, and we need to remove the overlap which is double-counted. So now, we make some conclusions: 1. the upper bound of the nth magical number should be n*b, where a is the smaller one (or b <= a); 2. there are n*b/a magical numbers smaller than n*b; 3. there are n*b/(minimum common multiple) overlaps. Thus, the overall count is: n + ...
Post a Comment
Read more

Segment Tree

Segment tree can be viewed as an abstract data structure which using some more space to trade for speed. For example, for a typical question with O(N^2) time complexity, the segment tree method can decrease it to O(N*log(N)).  To make it understandable, let us consider one example. Say we have an integer array of N size, and what we want is to query the maximum with a query range [idx1, idx2], where idx1 is the left indexes, and idx2 is the right indexes inclusive. If we only do this kind of query once, then we just need to scan through the array from idx1 to idx2 once, and record the maximum, done. The time complexity is O(N), which is decent enough in most cases even though it is not the optimal one (for example, with a segment tree built, the time complexity can decrease down to O(log(N))). However, how about we need to query the array N times? If we continue to use the naïve way above, then the time complexity is O(N^2), since for each query we need to scan the query range once...
Post a Comment
Read more

Recursion - Example

Recursion - Example Leetcode 231  Power of Two Given an integer n, return true if it is a power of two. Otherwise, return false. An integer n is a power of two, if there exists an integer x such that n == 2^x Constraints: -2^31 <= n <= 2^31 - 1 Analysis: One way is to think about this question recursively: if n%2 == 1, then n must not be power of 2; if not, then we just need to consider whether (n/2) is a power of 2 or not. This is exactly the "same question with a smaller size"! It is trivial to figure out the base cases: if n == 0, return false; if n == 1, return true. See the code below: class Solution { public: bool isPowerOfTwo(int n) { // base cases if(n == 0) return false; if(n == 1) return true; // converging if(n%2 == 1) return false; return isPowerOfTwo(n/2); } }; If interested, there are some other ways to solve this problem. For example, using bit manipulation, we can have the following solution: class ...
Post a Comment
Read more