Practice
1. You are given a triangular array of numbers where each element can only move to one of the two adjacent numbers in the row below. The goal is to find the minimum sum path from the top to the bottom. Which algorithmic approach guarantees an optimal solution efficiently?
easy
Solution
Step 1: Understand problem constraints and properties
The problem requires finding a minimum path sum with overlapping subproblems and optimal substructure, which suits dynamic programming.Step 2: Identify suitable algorithm
Bottom-up DP tabulation efficiently computes minimum sums from the bottom row up, avoiding recomputation and ensuring optimality.Final Answer:
Option D -> Option DQuick Check:
DP tabulation solves overlapping subproblems efficiently [OK]
Hint: DP bottom-up tabulation ensures optimal substructure [OK]
Common Mistakes:
- Assuming greedy choice is always optimal
- Using brute force recursion without memoization
- Trying to split triangle independently ignoring dependencies
2. Consider the following snippet from the optimal Zuma Game solution. Given
board = "WRRBBW" and hand = "RB", what is the return value of findMinStep(board, hand)?easy
Solution
Step 1: Trace initial calls with board="WRRBBW" and hand="RB"
Hand has 1 R and 1 B. The board requires more balls to clear groups of 3, but hand is insufficient to clear all.Step 2: Check if any insertion sequence clears board
Trying to insert R or B balls to form groups of 3 fails due to insufficient balls; no sequence clears the board fully.Final Answer:
Option B -> Option BQuick Check:
Return -1 indicates no solution with given hand [OK]
Hint: Insufficient hand balls cause no solution -> -1 [OK]
Common Mistakes:
- Assuming partial removals suffice to clear board
- Miscounting balls needed for group removal
- Forgetting to restore hand counts after recursion
3. The following code snippet is a near-correct implementation of the Cherry Pickup DP solution. Identify the line containing the subtle bug that can cause incorrect cherry counting or invalid states.
for step in range(1, 2 * (n - 1) + 1):
for r1 in range(max(0, step - (n - 1)), min(n, step + 1)):
c1 = step - r1
if c1 < 0 or c1 >= n:
continue
for r2 in range(max(0, step - (n - 1)), min(n, step + 1)):
c2 = step - r2
if c2 < 0 or c2 >= n:
continue
if grid[r1][c1] == -1 or grid[r2][c2] == -1:
continue
candidates = []
if r1 > 0 and r2 > 0:
candidates.append(dp[r1 - 1][c1][r2 - 1])
if r1 > 0 and c2 > 0:
candidates.append(dp[r1 - 1][c1][r2])
if c1 > 0 and r2 > 0:
candidates.append(dp[r1][c1 - 1][r2 - 1])
if c1 > 0 and c2 > 0:
candidates.append(dp[r1][c1 - 1][r2])
best_prev = max(candidates) if candidates else -float('inf')
if best_prev == -float('inf'):
continue
val = best_prev + grid[r1][c1]
if r1 != r2:
val += grid[r2][c2]
dp[r1][c1][r2] = max(dp[r1][c1][r2], val)
medium
Solution
Step 1: Examine candidate state transitions
The candidates appended correspond to previous dp states, but the code does not ensure both players have taken the same number of steps, leading to invalid states.Step 2: Identify the bug
Not verifying that the previous states correspond to the same step count causes incorrect cherry counting or index errors.Final Answer:
Option A -> Option AQuick Check:
Ensuring step synchronization is critical for correctness [OK]
Hint: Check if previous states align on step count [OK]
Common Mistakes:
- Double counting cherries when players overlap
- Ignoring thorn cells early
- Using 4D dp instead of 3D
4. What is the time complexity of the space-optimized bottom-up dynamic programming solution for the Unique Paths problem on an
m x n grid?medium
Solution
Step 1: Identify loops in the code
The solution uses two nested loops: outer loop runs m-1 times, inner loop runs n-1 times.Step 2: Calculate total operations
Total operations ≈ (m-1) * (n-1) -> O(m * n). No extra hidden loops or recursion stack.Final Answer:
Option D -> Option DQuick Check:
Nested loops over m and n -> O(m*n) [OK]
Hint: Nested loops over m and n -> O(m*n) [OK]
Common Mistakes:
- Confusing with recursion exponential time
- Forgetting loops multiply complexity
5. Suppose the Cherry Pickup problem is modified so that players can revisit cells multiple times (i.e., cycles allowed), and cherries are replenished after being picked (can be collected multiple times). Which of the following algorithmic changes correctly adapts the solution to this variant?
hard
Solution
Step 1: Understand the impact of allowing revisits and replenished cherries
Cycles and replenished cherries mean the problem is no longer acyclic and DP assumptions break down.Step 2: Identify suitable algorithm
Modeling the problem as a shortest path on a state graph with both players' positions and steps allows handling cycles and repeated cherry collection correctly.Final Answer:
Option D -> Option DQuick Check:
DP fails with cycles; shortest path algorithms handle repeated states [OK]
Hint: Cycles break DP; use graph shortest path [OK]
Common Mistakes:
- Removing DP constraints without changing algorithm
- Adding dimensions to DP without handling cycles
- Using greedy which ignores global optimality