LeetCode 453 | Minimum Moves to Equal Array Elements | Java | Medium | Math & Greedy | CodeByTushu
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In today’s video, we solve LeetCode 453 — Minimum Moves to Equal Array Elements using Math and Greedy invariants in Java.
We are given an integer array nums of size n, and in each move, we can increment n - 1 elements by 1. Our objective is to determine the minimum number of moves required to make all elements equal. While a naive simulation would be far too slow, transforming our perspective on the problem reveals that incrementing n - 1 elements is mathematically equivalent to decrementing 1 element relative to all others.
Under this relative difference invariant, every move simply reduces an element above the minimum closer to the baseline. Therefore, the minimum moves required to equalize all elements is simply the sum of differences between every array value and the global minimum element. We identify the minimum in a single pass and accumulate the differences using a 64-bit accumulator to eliminate integer overflow, achieving optimal O(N) time and O(1) auxiliary space.
🔥 Problem Highlights
✅ Mathematical transformation: incrementing (n - 1) elements is identical to decrementing 1 element
✅ Global minimum element serves as the optimal convergence target baseline
✅ Single-pass linear scan avoids costly sorting operations
✅ 64-bit integer moves accumulator prevents potential arithmetic overflow
✅ Minimal, branch-free arithmetic logic guarantees high CPU cache throughput
✅ Optimal O(N) linear runtime and O(1) strictly constant auxiliary memory
⏱️ TIMESTAMPS
00:00 Intro & November Kickoff (Day 99 Streak - 1 Day to Century)
00:39 Problem Understanding & Constraints Analysis
01:43 Key Breakthrough: Incrementing (N-1) == Decrementing 1
02:23 Optimal Linear Algorithm & Global Min Baseline
03:03 Clean Java 17 Solution Walkthrough Line by Line
03:41 Test Cases Walkthrough & Day 99 Streak Celebration
04:18 Time Complexity Analysis (O(N) Linear Scan)
04:46 Space Complexity Analysis (O(1) Constant Space)
05:08 Outro & Tomorrow Day 100 Century Milestone!
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/November/01-November/453-Minimum-Moves-to-Equal-Array-Elements
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/November/01-November/453-Minimum-Moves-to-Equal-Array-Elements/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/November/01-November/453-Minimum-Moves-to-Equal-Array-Elements/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/November/01-November/453-Minimum-Moves-to-Equal-Array-Elements/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/November/01-November/453-Minimum-Moves-to-Equal-Array-Elements/solution.js
🔗 LeetCode Source Code
📌 https://codebytushu.com/leetcode/problem/minimum-moves-to-equal-array-elements
🌐 All LeetCode Solutions
📌 https://codebytushu.com/Leetcode/
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🔥 Perfect For
LeetCode Medium Problems
Java DSA Preparation
Math & Relative Difference Invariants
Greedy Baseline Alignment
Array Transformation Optimization
Time Complexity Optimization
Coding Interview Preparation
Problem Solving
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💬 Question of the Day
Why is incrementing n - 1 elements by 1 mathematically equivalent to decrementing 1 element by 1 in this problem?
1️⃣ Because relative differences between elements are completely preserved
2️⃣ Because it alters the parity of the entire array uniformly
3️⃣ Because it forces the maximum value to divide evenly
4️⃣ Because it allows sorting in logarithmic time
#LeetCode453 #LeetCode #MinimumMovesToEqualArrayElements #Java #JavaDSA #Math #Greedy #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
In today’s video, we solve LeetCode 453 — Minimum Moves to Equal Array Elements using Math and Greedy invariants in Java.
We are given an integer array nums of size n, and in each move, we can increment n - 1 elements by 1. Our objective is to determine the minimum number of moves required to make all elements equal. While a naive simulation would be far too slow, transforming our perspective on the problem reveals that incrementing n - 1 elements is mathematically equivalent to decrementing 1 element relative to all others.
Under this relative difference invariant, every move simply reduces an element above the minimum closer to the baseline. Therefore, the minimum moves required to equalize all elements is simply the sum of differences between every array value and the global minimum element. We identify the minimum in a single pass and accumulate the differences using a 64-bit accumulator to eliminate integer overflow, achieving optimal O(N) time and O(1) auxiliary space.
🔥 Problem Highlights
✅ Mathematical transformation: incrementing (n - 1) elements is identical to decrementing 1 element
✅ Global minimum element serves as the optimal convergence target baseline
✅ Single-pass linear scan avoids costly sorting operations
✅ 64-bit integer moves accumulator prevents potential arithmetic overflow
✅ Minimal, branch-free arithmetic logic guarantees high CPU cache throughput
✅ Optimal O(N) linear runtime and O(1) strictly constant auxiliary memory
⏱️ TIMESTAMPS
00:00 Intro & November Kickoff (Day 99 Streak - 1 Day to Century)
00:39 Problem Understanding & Constraints Analysis
01:43 Key Breakthrough: Incrementing (N-1) == Decrementing 1
02:23 Optimal Linear Algorithm & Global Min Baseline
03:03 Clean Java 17 Solution Walkthrough Line by Line
03:41 Test Cases Walkthrough & Day 99 Streak Celebration
04:18 Time Complexity Analysis (O(N) Linear Scan)
04:46 Space Complexity Analysis (O(1) Constant Space)
05:08 Outro & Tomorrow Day 100 Century Milestone!
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/November/01-November/453-Minimum-Moves-to-Equal-Array-Elements
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/November/01-November/453-Minimum-Moves-to-Equal-Array-Elements/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/November/01-November/453-Minimum-Moves-to-Equal-Array-Elements/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/November/01-November/453-Minimum-Moves-to-Equal-Array-Elements/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/November/01-November/453-Minimum-Moves-to-Equal-Array-Elements/solution.js
🔗 LeetCode Source Code
📌 https://codebytushu.com/leetcode/problem/minimum-moves-to-equal-array-elements
🌐 All LeetCode Solutions
📌 https://codebytushu.com/Leetcode/
📱 Join CodeByTushu on WhatsApp
Get coding updates, LeetCode solutions, daily questions and more.
https://whatsapp.com/channel/0029Vb8k5mCDeON2Xwlkxe40
❤️ Donate
📌 https://codebytushu.com/Leetcode/donate.php
🔥 Perfect For
LeetCode Medium Problems
Java DSA Preparation
Math & Relative Difference Invariants
Greedy Baseline Alignment
Array Transformation Optimization
Time Complexity Optimization
Coding Interview Preparation
Problem Solving
🌐 Follow CodeByTushu
▶️ YouTube: https://www.youtube.com/@codebytushu
📸 Instagram: https://instagram.com/codebytushu/
💼 LinkedIn: https://www.linkedin.com/company/codebytushu/
🌍 Website: https://codebytushu.com
🐙 GitHub: https://github.com/CodeByTushu
❤️ Don’t Forget To
👍 Like | 💬 Comment | 🔔 Subscribe
💬 Question of the Day
Why is incrementing n - 1 elements by 1 mathematically equivalent to decrementing 1 element by 1 in this problem?
1️⃣ Because relative differences between elements are completely preserved
2️⃣ Because it alters the parity of the entire array uniformly
3️⃣ Because it forces the maximum value to divide evenly
4️⃣ Because it allows sorting in logarithmic time
#LeetCode453 #LeetCode #MinimumMovesToEqualArrayElements #Java #JavaDSA #Math #Greedy #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description