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Maximum Array Sum After B Negations

Logical Breakdown

  • [ ] Subproblem Identification:
  • [ ] Optimal Substructure:
  • [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
  • [ ] Optimization: (Matrix Exponentiation if 'A' is huge)

Mathematical Rigor

State Definition

Let \(dp[i]\) be the state for the \(i\)-th subproblem.

Recurrence Relation

\[ dp[i] = ... \]

Visualization

graph TD A["Current State"] --> B["Previous State 1"] A --> C["Previous State 2"]

Complexity Analysis

Approach Time Complexity Space Complexity
Iterative DP \(O(N)\) \(O(1)\)

Code Reference

package com.dsa.heaps;

/* * ============================================================================================ * PROBLEM: Maximum Array Sum After B Negations * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given an array A and an integer B. You must perform exactly B negations (A[i] = -A[i]). * Goal: Maximize the total sum of the array. * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * To maximize the sum, we must follow a greedy strategy: * * Step 1: Greedy Choice * - In each step, we should negate the smallest element currently in the array. * - If the smallest element is negative, negating it makes it positive, providing the * largest possible increase to the sum. * - If all elements are positive, negating the smallest positive number results in * the smallest possible decrease to the sum. * * Step 2: Efficiency via Min-Heap * - Since we need the minimum element repeatedly across B operations, a Min-Heap * is optimal. * * Algorithm: * 1. Build a Min-Heap from all elements in array A. * 2. Repeat B times: * a. Extract the minimum element minVal. * b. Negate it: newVal = -minVal. * c. Insert newVal back into the Min-Heap. * 3. After B operations, extract all elements from the heap and sum them up. * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Constraints: * 1 <= N <= 5 * 10^5, 1 <= B <= 5 * 10^6 * * Time Complexity: * O(N + B log N) * - Heapification: O(N). * - B operations of extract-min and insert: O(B log N). * * Space Complexity: * O(N) to store the elements in the Priority Queue. * ============================================================================================ / public class MaximumArraySumAfterBNegations { public int solve(int[] a4, int A) { // TODO: Implement solution return 0; } }