Distribute Candy
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.heaps;
import java.util.*;
/*
* ============================================================================================
* PROBLEM: Distribute Candy (Greedy Neighbors)
* ============================================================================================
*
* ---------------------
* 1. Logical Breakdown & Mathematical Formulation
* ---------------------
* We must satisfy two conditions for every child i:
* 1. \(C[i] \ge 1\) (At least one candy)
* 2. If \(Rating[i] > Rating[i-1]\), then \(C[i] > C[i-1]\)
* 3. If \(Rating[i] > Rating[i+1]\), then \(C[i] > C[i+1]\)
*
* Step 1: Left-to-Right Pass
* - Initialize an array candies with 1 for all children.
* - Iterate from 1 to N-1:
* - If \(A[i] > A[i-1]\), set \(candies[i] = candies[i-1] + 1\).
* - This ensures every child has more candies than their left neighbor if their
* rating is higher.
*
* Step 2: Right-to-Left Pass
* - Iterate from N-2 down to 0:
* - If \(A[i] > A[i+1]\), the child needs more candies than the right neighbor.
* - To satisfy both passes, set \(candies[i] = max(current\_candies[i], candies[i+1] + 1)\).
*
* Step 3: Result
* - The sum of the candies array is the minimum total candies required.
*
* ---------------------
* 2. Complexity Analysis
* ---------------------
* Time Complexity:
* O(N)
* - Two independent linear passes over the rating array.
*
* Space Complexity:
* O(N)
* - To store the candies count for each child.
*
* ---------------------
* 3. Base Case Verification
* ---------------------
* - N=1: Only 1 child, returns 1 candy.
* - All ratings equal: Each child gets 1 candy, total = N.
* - Strictly increasing ratings: Candies will be [1, 2, 3, ..., N].
* ============================================================================================
/
public class DistributeCandy {
public int solve(int A) {
// TODO: Implement solution
return 0;
}
public int candy(int[] a3) {
return 0;
}
}