Sum The Difference
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
Visualization
Complexity Analysis
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Iterative DP | \(O(N)\) | \(O(1)\) |
Code Reference
package com.dsa.recursion;
/ * Problem: Sum The Difference * Group: 07. Recursion */ / * ============================================================================================ * PROBLEM: Sum the Difference * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given an integer array A of size N. * Task: * 1. Find all possible non-empty subsequences. * 2. For each subsequence, calculate (Max_Value - Min_Value). * 3. Return the sum of all these differences modulo 10^9 + 7. * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * Naive Approach: * Generating all subsequences takes O(2^N). For N=10,000, 2^10000 is impossibly large. * We must use the "Contribution Technique". * * Equation: * Total Sum = Sum(Max(subsequence) - Min(subsequence)) * = Sum(Max(subsequence)) - Sum(Min(subsequence)) * * Strategy: * 1. Sort the array A in ascending order. * 2. For each element A[i], calculate how many subsequences have A[i] as the Maximum. * 3. For each element A[i], calculate how many subsequences have A[i] as the Minimum. * * Calculating Max Contribution: * - If A is sorted, A[i] is the maximum of a subsequence if the subsequence consists of A[i] * and any subset of the elements smaller than A[i] (indices 0 to i-1). * - There are 'i' elements smaller than A[i]. * - Number of such subsequences = 2^i. * - Contribution to Max Sum = A[i] * 2^i. * * Calculating Min Contribution: * - A[i] is the minimum of a subsequence if the subsequence consists of A[i] * and any subset of the elements larger than A[i] (indices i+1 to N-1). * - There are 'N - 1 - i' elements larger than A[i]. * - Number of such subsequences = 2^(N - 1 - i). * - Contribution to Min Sum = A[i] * 2^(N - 1 - i). * * Final Formula: * Result = Sum [ (A[i] * 2^i) - (A[i] * 2^(N - 1 - i)) ] * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Constraints: * 1 <= N <= 10,000 * 1 <= A[i] <= 1,000 * * Time Complexity: * O(N log N) - Dominated by the sorting step. * The linear pass to calculate powers and sums is O(N). * * Space Complexity: * O(N) or O(1) - Depending on whether we precompute powers of 2. * * Critical Implementation Detail (Modulo Arithmetic): * The result involves subtraction: (MaxSum - MinSum). * In modular arithmetic, (A - B) % M can be negative. * Fix: ((MaxSum - MinSum) % M + M) % M. * * --------------------- * 4. Input / Output Format * --------------------- * Input: Integer array A. * Output: Integer result modulo 10^9 + 7. * * --------------------- * 5. Examples * --------------------- * Input 1: A = [1, 2] * Sorted: [1, 2] * i=0 (Val 1): Max Contrib = 1 * 2^0 = 1. Min Contrib = 1 * 2^1 = 2. * i=1 (Val 2): Max Contrib = 2 * 2^1 = 4. Min Contrib = 2 * 2^0 = 2. * Total Max = 1+4=5. Total Min = 2+2=4. Result = 5-4 = 1. * * Input 2: A = [3, 5, 10] * Sorted: [3, 5, 10] * Powers of 2: [1, 2, 4] * Max Sum: (31) + (52) + (104) = 3 + 10 + 40 = 53 * Min Sum: (34) + (52) + (101) = 12 + 10 + 10 = 32 * Result: 53 - 32 = 21. * * ============================================================================================ */ public class SumTheDifference { public int solve(int[] A) { // TODO: Implement solution return 0; } }