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Minimum One Bit Operations To Make Integers Zero - Leetcode Solution

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LeetCode:  Minimum One Bit Operations To Make Integers Zero Leetcode Solution

Difficulty: Hard

Topics: dynamic-programming bit-manipulation  

Problem Statement:

Given an array A of integers, we must modify the array in the following way: we choose an i and replace A[i] with -A[i], and we repeat this process K times in total. (We may choose the same index i multiple times.)

Return the largest possible sum of the array after modifying it in this way.

Example 1: Input: A = [4,2,3], K = 1 Output: 5 Explanation: Choose indices (1,) and A becomes [4,-2,3].

Example 2: Input: A = [3,-1,0,2], K = 3 Output: 6 Explanation: Choose indices (1, 2, 2) and A becomes [3,1,0,2].

Solution:

To solve this problem, we need to understand the bit manipulation operations. The two most important operations are bitwise AND and bitwise XOR.

Bitwise AND:

This operation returns a value by comparing each corresponding bit of the operands. If the corresponding bits of both operands are 1, it returns 1. Otherwise, it returns 0.

Bitwise XOR:

This operation returns a value by comparing each corresponding bit of the operands. If the corresponding bits of both operands are different, it returns 1. Otherwise, it returns 0.

If we perform bitwise XOR with a number and its two's complement, we get all 1's. If we perform a bitwise AND with a number and its two's complement, we get all 0's. These properties are used to manipulate bits efficiently.

To solve the given problem, we can use the following steps:

Step 1: Find the maximum absolute value in the array A. Step 2: If K is even, return the sum of all elements of A. Step 3: If K is odd, replace the maximum absolute value with its negation and return the sum of all elements of A.

Let's understand these steps with an example:

Example: A = [4, 2, 3], K = 1

Step 1: The maximum absolute value is 4. Step 2: K is odd, so we replace 4 with -4. The array becomes [-4, 2, 3]. The sum of the array is 1. Step 3: Return 1.

In the above example, we need to perform one bit operation to make the array sum maximum. Similarly, we can perform K bit operations to make the array sum maximum.

Code:

The code to solve the given problem is as follows:

class Solution { public: int largestSumAfterKNegations(vector<int>& A, int K) { int sum = 0; int max_abs = INT_MIN; for (int i = 0; i < A.size(); i++) { sum += A[i]; max_abs = max(max_abs, abs(A[i])); } if (K % 2 == 0) return sum; else return sum - 2 * max_abs; } };

The above code uses the in-built function max() to find the maximum absolute value in the array A. It then performs the operations as described above, to return the maximum sum of the array after performing K bit operations.

Minimum One Bit Operations To Make Integers Zero Solution Code

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