Practice
1. You are given a 2D binary matrix filled with 0's and 1's. The task is to find the largest square containing only 1's and return its area. Which algorithmic approach guarantees an optimal solution with polynomial time complexity?
easy
Solution
Step 1: Understand problem constraints
The problem requires finding the largest square of 1's, which depends on overlapping subproblems of smaller squares.Step 2: Recognize DP pattern
Dynamic Programming efficiently computes maximal squares by using previously computed smaller squares, avoiding redundant checks.Final Answer:
Option D -> Option DQuick Check:
DP uses subproblem solutions to build larger squares [OK]
Hint: Largest square -> DP using smaller squares [OK]
Common Mistakes:
- Thinking greedy expansion always works
- Confusing DFS with DP
- Assuming sorting helps find squares
2. You are given an
n x n integer matrix. You want to find a path from the top row to the bottom row such that you move one step down each time, and at each step you can move to the same column, the column to the left, or the column to the right. The goal is to minimize the sum of the values along this path. Which algorithmic approach guarantees finding the minimum sum efficiently?easy
Solution
Step 1: Understand the problem constraints
The problem requires considering all possible paths moving down and diagonally, which suggests overlapping subproblems and optimal substructure.Step 2: Identify the suitable algorithm
Dynamic programming fits because it efficiently computes minimum sums for each cell based on the previous row's results, avoiding redundant calculations.Final Answer:
Option D -> Option DQuick Check:
Greedy and sorting fail to consider future steps; DFS without memoization is exponential [OK]
Hint: DP uses previous row results to build solutions [OK]
Common Mistakes:
- Thinking greedy or sorting per row suffices
- Ignoring overlapping subproblems
3. You need to find the number of distinct ways to move from the top-left corner to the bottom-right corner of an
m x n grid, moving only down or right at each step. Which algorithmic approach guarantees an efficient and optimal solution for this problem?easy
Solution
Step 1: Understand problem constraints
The problem requires counting all unique paths with only right and down moves, which naturally forms overlapping subproblems.Step 2: Identify suitable approach
Dynamic Programming efficiently solves overlapping subproblems by storing intermediate results, unlike greedy or pure recursion which are either incorrect or inefficient.Final Answer:
Option C -> Option CQuick Check:
DP uses subproblem reuse -> optimal and efficient [OK]
Hint: Counting paths with overlapping subproblems -> DP [OK]
Common Mistakes:
- Thinking greedy can find all paths
- Using pure recursion without memoization
4. What is the time complexity of the space-optimized bottom-up DP solution for the Cheapest Flights Within K Stops problem, given n cities, E flights, and maximum K stops?
medium
Solution
Step 1: Identify loops in the algorithm
The outer loop runs K+1 times, and the inner loop iterates over all E flights to relax edges.Step 2: Calculate total complexity
Each iteration processes E edges, so total time is O(K * E). No nested loops over n^2 or log factors appear.Final Answer:
Option A -> Option AQuick Check:
Algorithm iterates over edges K+1 times [OK]
Hint: Outer loop K+1 times, inner loop over E edges [OK]
Common Mistakes:
- Confusing E with n^2
- Assuming Dijkstra complexity
- Counting recursion stack space
5. The following code attempts to solve the Stone Game problem using bottom-up DP. Identify the line containing the subtle bug that causes incorrect results on some inputs.
def stoneGame(piles):
n = len(piles)
dp = [[0]*n for _ in range(n)]
for i in range(n):
dp[i][i] = piles[i]
for length in range(2, n+1):
for i in range(n - length + 1):
j = i + length - 1
dp[i][j] = max(piles[i] + dp[i+1][j], piles[j] + dp[i][j-1])
return dp[0][n-1] > 0
medium
Solution
Step 1: Understand dp state meaning
dp[i][j] should represent the maximum difference in stones the current player can achieve over the opponent.Step 2: Identify incorrect recurrence
The buggy line adds piles[i] + dp[i+1][j], which ignores the opponent's optimal response. The correct formula subtracts dp[i+1][j] to account for opponent's best play.Final Answer:
Option A -> Option AQuick Check:
Correct recurrence uses subtraction, not addition [OK]
Hint: DP recurrence must subtract opponent's score [OK]
Common Mistakes:
- Adding dp values instead of subtracting opponent's score
- Forgetting base case initialization