Max Sum Without Adjacent Elements
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.dynamic_programming;
import java.util.*;
/ * Problem: Max Sum Without Adjacent Elements * Group: 11. Dynamic Programming */ / * ============================================================================================ * PROBLEM: Max Sum Without Adjacent Elements * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given a 2 x N grid of integers A. * Choose numbers to maximize sum such that no two numbers are adjacent. * * --------------------- * 2. Logical Breakdown * --------------------- * Key Observation: * Since we cannot pick adjacent elements, for any column 'i', we can pick at most ONE element. * (If we pick both top and bottom of column 'i', they are vertically adjacent). * * Simplification: * Let V[i] = max(A[0][i], A[1][i]). * The problem reduces to finding the max sum in array V such that no two elements are adjacent. * * Recurrence Relation (House Robber style): * Let dp[i] be the max sum considering columns 0 to i. * At column 'i', we have two choices: * 1. Exclude V[i]: Sum = dp[i-1] * 2. Include V[i]: Sum = V[i] + dp[i-2] (Must skip i-1) * * Formula: dp[i] = max(dp[i-1], V[i] + dp[i-2]) * * --------------------- * 3. Constraints & Complexity * --------------------- * Time: O(N) * Space: O(N) or O(1) optimized. * * ============================================================================================ */ public class MaxSumWithoutAdjacentElements { public int solve(int A) { // TODO: Implement solution return 0; }
public int adjacent(int[][] input1) {
return 1;
}
}