What is the time complexity of Minimize the Difference Between Target and Chosen Elements?

The optimal solution has a time complexity of O(m * n * min(5000, target)).

What is the space complexity of Minimize the Difference Between Target and Chosen Elements?

The space complexity is O(min(5000, target)).

What data structures are used to solve Minimize the Difference Between Target and Chosen Elements?

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 Minimize the Difference Between Target and Chosen Elements?

Insight 1: Dynamic Programming: We can use dynamic programming to build up the possible sums iteratively. We start with an initial sum of 0 and for each row, we update the possible sums by adding each element in the row to the existing possible sums. Insight 2: Using Sets for Efficiency: Using sets to store the possible sums avoids redundant calculations and duplicate sums, leading to better performance.

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Minimize the Difference Between Target and Chosen Elements - LeetCode 1981 Solution

Q: Which companies ask LeetCode 1981. Minimize the Difference Between Target and Chosen Elements in interviews?

This problem is frequently asked at Deutsche Bank, Sprinklr.

Minimize the Difference Between Target and Chosen Elements - Complete Solution Guide

Minimize the Difference Between Target and Chosen Elements is LeetCode problem 1981, 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 an m x n integer matrix mat and an integer target . Choose one integer from each row in the matrix such that the absolute difference between target and the sum of the chosen elements is minimized . Return the minimum absolute difference . The absolute difference between two numbers a and b is the absolute value of a - b . Example 1: Input: mat = [[1,2,3],[4,5,6],[7,8,9]], target = 13 Output: 0 Explanation: One possible choice is to: - Choose 1 from the first row. - Choose 5 from th

Detailed Explanation

The problem requires us to find the minimum absolute difference between a target value and the sum of chosen elements from a matrix. We must select exactly one integer from each row of the matrix. The goal is to minimize `abs(target - sum)`, where `sum` is the sum of the chosen integers.

Solution Approach

The solution employs a dynamic programming approach. It maintains a set of possible sums that can be achieved by selecting one element from each row processed so far. For each row, it iterates through the current set of possible sums and the unique elements in the current row. It calculates the new sums by adding an element from the current row to each existing possible sum. If a new sum is greater than or equal to the target, it only keeps track of the minimum of these 'above target' sums to avoid the set of possible sums growing unnecessarily large. Finally, after processing all rows, it calculates the absolute difference between each possible sum and the target, and returns the minimum of these differences.

Step-by-Step Algorithm

Step 1: Initialize a set `possible_sums` with the value 0. This set will store all possible sums achievable so far.
Step 2: Iterate through each row of the matrix.
Step 3: For each row, create a new set `next_sums` to store the possible sums achievable after including the current row. Also initialize `min_above` to infinity to keep track of minimum value that is greater than or equal to the target
Step 4: Iterate through the `possible_sums` and the unique elements of the current row.
Step 5: For each combination of `s` in `possible_sums` and `x` in the current row, calculate `new_sum = s + x`.
Step 6: If `new_sum` is less than `target`, add `new_sum` to `next_sums`. Otherwise, update `min_above` to the minimum value if `new_sum` is smaller than current `min_above`.
Step 7: After processing the current row, if `min_above` is not infinity, then include the value `min_above` to `next_sums` to minimize the space
Step 8: Update `possible_sums` to `next_sums` and repeat from step 2
Step 9: After processing all rows, iterate through the final `possible_sums` and find the minimum absolute difference between each sum and the `target`. Return the result.

Key Insights

Insight 1: Dynamic Programming: We can use dynamic programming to build up the possible sums iteratively. We start with an initial sum of 0 and for each row, we update the possible sums by adding each element in the row to the existing possible sums.
Insight 2: Using Sets for Efficiency: Using sets to store the possible sums avoids redundant calculations and duplicate sums, leading to better performance.
Insight 3: Early Termination for Sums Above Target: The provided solutions optimize by tracking the minimum sum that exceeds the target. This is because sums far above the target will likely result in larger differences, so only the closest one is retained. While the problem description doesn't explicitly set an upper bound for intermediate sums, the constraints imply a practical limit due to the constraint `1 <= target <= 800`.

Complexity Analysis

Time Complexity: O(m * n * min(5000, target))

Space Complexity: O(min(5000, target))

Topics

This problem involves: Array, Dynamic Programming, Matrix.

Companies

Asked at: Deutsche Bank, Sprinklr.

Minimize the Difference Between Target and Chosen Elements | LeetCode 1981

Medium

Array Dynamic Programming Matrix

Time: O(m * n * min(5000, target))Space: O(min(5000, target))

Solution Code

Available in:

import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.Arrays;

class Solution {
    public int minimizeTheDifference(int[][] mat, int target) {
        Set<Integer> possibleSums = new HashSet<>();
        possibleSums.add(0);

        for (int[] row : mat) {
            Set<Integer> nextSums = new HashSet<>();
            int minAbove = Integer.MAX_VALUE;

            Set<Integer> uniqueElements = new HashSet<>();
            for (int x : row) {
                uniqueElements.add(x);
            }

            for (int s : possibleSums) {
                for (int x : uniqueElements) {
                    int newSum = s + x;
                    if (newSum < target) {
                        nextSums.add(newSum);
                    } else {
                        minAbove = Math.min(minAbove, newSum);
                    }
                }
            }

            if (minAbove != Integer.MAX_VALUE) {
                nextSums.add(minAbove);
            }

            possibleSums = nextSums;
        }

        int minDifference = Integer.MAX_VALUE;
        for (int s : possibleSums) {
            minDifference = Math.min(minDifference, Math.abs(s - target));
        }

        return minDifference;
    }
}

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Problem Info

Description

Problem 1981: Minimize the Difference Between Target and Chosen Elements - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix

Companies

Deutsche BankSprinklr

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1981

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

9 steps

1
Step 1: Initialize a set `possible_sums` with the value 0. This set will store all possible sums achievable so far.
2
Step 2: Iterate through each row of the matrix.
3
Step 3: For each row, create a new set `next_sums` to store the possible sums achievable after including the current row. Also initialize `min_above` to infinity to keep track of minimum value that is greater than or equal to the target
4
Step 4: Iterate through the `possible_sums` and the unique elements of the current row.
5
Step 5: For each combination of `s` in `possible_sums` and `x` in the current row, calculate `new_sum = s + x`.
6
Step 6: If `new_sum` is less than `target`, add `new_sum` to `next_sums`. Otherwise, update `min_above` to the minimum value if `new_sum` is smaller than current `min_above`.
7
Step 7: After processing the current row, if `min_above` is not infinity, then include the value `min_above` to `next_sums` to minimize the space
8
Step 8: Update `possible_sums` to `next_sums` and repeat from step 2
9
Step 9: After processing all rows, iterate through the final `possible_sums` and find the minimum absolute difference between each sum and the `target`. Return 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 * min(5000, target))

💾

Space Complexity

O(min(5000, target))

📝

Why?

The outer loop iterates `m` times (number of rows). The inner loops iterate over the possible sums (at most `min(sum_limit, target)`) and the unique elements in each row (at most `n`). The space complexity comes from storing the `possible_sums` set, which can contain at most `min(sum_limit, target)` elements.

📖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

Generate all possible combinations of selecting one element from each row, calculate the sum for each combination, and then find the minimum absolute difference. However, this approach has exponential time complexity.

Time: O(n^m), where n is the number of columns and m is the number of rows.Space: O(m), for storing the current combination of selected elements.

Trade-offs: The brute force approach is simple to understand but highly inefficient. It is only feasible for very small input sizes (very small m and n). The dynamic programming approach is much more efficient for larger inputs.

✨

Key Insights

1Insight 1: Dynamic Programming: We can use dynamic programming to build up the possible sums iteratively. We start with an initial sum of 0 and for each row, we update the possible sums by adding each element in the row to the existing possible sums.
2Insight 2: Using Sets for Efficiency: Using sets to store the possible sums avoids redundant calculations and duplicate sums, leading to better performance.
3Insight 3: Early Termination for Sums Above Target: The provided solutions optimize by tracking the minimum sum that exceeds the target. This is because sums far above the target will likely result in larger differences, so only the closest one is retained. While the problem description doesn't explicitly set an upper bound for intermediate sums, the constraints imply a practical limit due to the constraint `1 <= target <= 800`.

📚

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: mat = [[1]], target = 100. The algorithm will correctly return 99.
Edge case 2: Large values in the matrix and large target: The set `possible_sums` might grow very large, but the `min_above` optimization helps to mitigate this. The explicit constraints `1 <= mat[i][j] <= 70` and `1 <= target <= 800` help to keep the number of sums manageable.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a set to store `possible_sums`. This will lead to duplicate calculations and a much larger set, increasing both time and space complexity significantly.

✗Common mistake 2: Not considering the optimization of only storing the minimum sum greater than or equal to the target. This can lead to `possible_sums` becoming very large and inefficient.

✅

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: Portfolio optimization: Given a set of investments with varying returns and risks, select one investment from each category to minimize the difference between the desired return and the actual return.

🏦Real-world scenario 2: Resource allocation: Given a set of tasks and multiple ways to perform each task with varying costs, select one way to perform each task to minimize the difference between the desired budget and the actual cost.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: June 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimize the Difference Between Target and Chosen Elements"?

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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