What is the time complexity of Minimum Number of Operations to Satisfy Conditions?

The optimal solution has a time complexity of O(m*n).

What data structures are used to solve Minimum Number of Operations to Satisfy Conditions?

This problem uses Array, Dynamic Programming, Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Number of Operations to Satisfy Conditions?

Insight 1: The problem can be solved column by column using dynamic programming. The state represents the cost of making the current column have a certain value. Insight 2: The previous column only impacts the current column through a minimum cost constraint. Since grid values are small (0-9), we can iterate through all possible values efficiently.

Which companies ask LeetCode 3122. Minimum Number of Operations to Satisfy Conditions in interviews?

This problem is frequently asked at Turing.

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Minimum Number of Operations to Satisfy Conditions - LeetCode 3122 Solution

Minimum Number of Operations to Satisfy Conditions - Complete Solution Guide

Minimum Number of Operations to Satisfy Conditions is LeetCode problem 3122, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 2D matrix grid of size m x n . In one operation , you can change the value of any cell to any non-negative number. You need to perform some operations such that each cell grid[i][j] is: Equal to the cell below it, i.e. grid[i][j] == grid[i + 1][j] (if it exists). Different from the cell to its right, i.e. grid[i][j] != grid[i][j + 1] (if it exists). Return the minimum number of operations needed. Example 1: Input: grid = [[1,0,2],[1,0,2]] Output: 0 Explanation: All the cells in t

Detailed Explanation

The problem requires finding the minimum number of operations to transform a given m x n grid such that each cell is equal to the cell directly below it (if it exists) and different from the cell immediately to its right (if it exists). An operation consists of changing the value of any cell to any non-negative number. The grid elements are integers between 0 and 9 inclusive.

Solution Approach

The solution uses dynamic programming to iterate through the columns of the grid. `dp[v]` represents the minimum cost to make all elements in the current column have value `v`. For each column, we calculate the cost to change the values in the current column to each possible value from 0 to 9. Then, we determine the minimum cost from the previous column to arrive at each possible value in the current column, considering the constraint that adjacent values should not be equal. The two smallest values of the previous column are used to handle the differing adjacency requirement.

Step-by-Step Algorithm

Step 1: Initialize `counts` array to store the frequency of each digit (0-9) in each column. This avoids redundant calculations.
Step 2: Initialize `dp` array of size 10. For the first column, `dp[v]` is initialized to the number of elements in the first column that are not equal to `v`.
Step 3: Iterate through the remaining columns (from column 1 to n-1).
Step 4: In each column, find the two minimum costs (`min1` and `min2`) from the previous column's `dp` array and also store the value (`min1_val`) associated with the first minimum cost.
Step 5: For each digit `v_curr` (0-9) in the current column, calculate the cost to change all elements in the current column to `v_curr`.
Step 6: Update `dp[v_curr]` by adding the cost to change the current column to `v_curr` to the minimum cost of the previous column. If `v_curr` is equal to the value associated with `min1`, then `min2` is used. Otherwise, `min1` is used.
Step 7: After iterating through all columns, the minimum value in the final `dp` array is the result.

Key Insights

Insight 1: The problem can be solved column by column using dynamic programming. The state represents the cost of making the current column have a certain value.
Insight 2: The previous column only impacts the current column through a minimum cost constraint. Since grid values are small (0-9), we can iterate through all possible values efficiently.
Insight 3: Finding the two smallest costs from the previous column's DP table allows us to efficiently handle the constraint that adjacent columns must have different values.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Matrix.

Companies

Asked at: Turing.

Minimum Number of Operations to Satisfy Conditions | LeetCode 3122

Medium

Array Dynamic Programming Matrix

Time: O(m*n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int minimumOperations(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        
        // Pre-calculate counts of each number for each column
        int[][] counts = new int[n][10];
        for (int j = 0; j < n; j++) {
            for (int i = 0; i < m; i++) {
                counts[j][grid[i][j]]++;
            }
        }
            
        // dp[v] will store the minimum cost for the current column to be value v
        int[] dp = new int[10];
        
        // Base case: first column (j=0)
        for (int v = 0; v < 10; v++) {
            dp[v] = m - counts[0][v];
        }
            
        // Iterate through the rest of the columns
        for (int j = 1; j < n; j++) {
            // Find the two minimum costs from the previous column's dp table
            int min1 = Integer.MAX_VALUE;
            int min2 = Integer.MAX_VALUE;
            int min1_val = -1;
            
            for (int v_prev = 0; v_prev < 10; v_prev++) {
                int cost_prev = dp[v_prev];
                if (cost_prev < min1) {
                    min2 = min1;
                    min1 = cost_prev;
                    min1_val = v_prev;
                } else if (cost_prev < min2) {
                    min2 = cost_prev;
                }
            }

            int[] new_dp = new int[10];
            for (int v_curr = 0; v_curr < 10; v_curr++) {
                int cost_to_change = m - counts[j][v_curr];
                if (v_curr == min1_val) {
                    new_dp[v_curr] = cost_to_change + min2;
                } else {
                    new_dp[v_curr] = cost_to_change + min1;
                }
            }
            dp = new_dp;
        }
            
        int minCost = Integer.MAX_VALUE;
        for (int cost : dp) {
            minCost = Math.min(minCost, cost);
        }
        return minCost;
    }
}

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Problem Info

Description

Problem 3122: Minimum Number of Operations to Satisfy Conditions - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix

Companies

Turing

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3122

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize `counts` array to store the frequency of each digit (0-9) in each column. This avoids redundant calculations.
2
Step 2: Initialize `dp` array of size 10. For the first column, `dp[v]` is initialized to the number of elements in the first column that are not equal to `v`.
3
Step 3: Iterate through the remaining columns (from column 1 to n-1).
4
Step 4: In each column, find the two minimum costs (`min1` and `min2`) from the previous column's `dp` array and also store the value (`min1_val`) associated with the first minimum cost.
5
Step 5: For each digit `v_curr` (0-9) in the current column, calculate the cost to change all elements in the current column to `v_curr`.
6
Step 6: Update `dp[v_curr]` by adding the cost to change the current column to `v_curr` to the minimum cost of the previous column. If `v_curr` is equal to the value associated with `min1`, then `min2` is used. Otherwise, `min1` is used.
7
Step 7: After iterating through all columns, the minimum value in the final `dp` array is the result.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested loops iterating through the grid's dimensions (m x n). The frequency count calculation takes O(m*n). The outer loop for the DP iteration takes O(n), and the inner loops to find the minimums and calculate the new DP values take O(1) time (since they only iterate through 10 values at most). The space complexity is dominated by the `counts` and `dp` arrays, which take space proportional to the number of columns (n) and the range of possible values (10, which is constant), respectively.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Trying every possible combination of values for each cell and checking if the conditions are satisfied. This is highly inefficient.

Time: O(10^(m*n)), where m is the number of rows and n is the number of columns.Space: O(m*n)

Trade-offs: This approach is impractical for any reasonably sized grid due to its exponential time complexity. It's only useful for extremely small grids for understanding the problem.

✨

Key Insights

1Insight 1: The problem can be solved column by column using dynamic programming. The state represents the cost of making the current column have a certain value.
2Insight 2: The previous column only impacts the current column through a minimum cost constraint. Since grid values are small (0-9), we can iterate through all possible values efficiently.
3Insight 3: Finding the two smallest costs from the previous column's DP table allows us to efficiently handle the constraint that adjacent columns must have different values.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: A grid with only one row. The code correctly handles this since the outer loop condition `j < n` will still function correctly, computing correct costs.
Edge case 2: A grid with only one column. The algorithm correctly iterates only once. The base case initializes the `dp` array with correct counts.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the case where the minimum cost value in the previous column is the same as the current value being evaluated. In this case, the second minimum cost must be used.

✗Common mistake 2: Incorrectly calculating the cost to change the current column to a specific value. The cost is the number of cells that are *not* already equal to that value.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression: Optimizing a dataset by minimizing changes needed to satisfy certain constraints or patterns, similar to optimizing storage or transmission.

🏦Real-world scenario 2: Image processing: Cleaning up an image by minimizing pixel changes while enforcing smoothness or other desired properties between neighboring pixels.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 2, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Number of Operations to Satisfy Conditions"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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