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LeetCode 1217 | Minimum Cost to Move Chips to The Same Position | Easy Problem | C++ Solution
In this video, I explain the problem step-by-step, discuss the optimal greedy approach, and implement the solution live while coding.
🚀 Problem: LeetCode 1217 – Minimum Cost to Move Chips to The Same Position
🧠 Approach:
Observe the movement cost rules:
Moving a chip by 2 positions costs 0
Moving a chip by 1 position costs 1
Count:
Number of chips at even positions
Number of chips at odd positions
To make all chips meet at the same position:
Move the smaller group to the larger group
Return the minimum of:
even count
odd count
⏱ Time Complexity: O(n)
📦 Space Complexity: O(1)
🔗 GitHub Repository:
https://github.com/VishwaPatel892/Leetcode_Solutions
👨💻 LeetCode Profile:
https://leetcode.com/u/vishwa0102/
💼 LinkedIn:
https://www.linkedin.com/in/vishwa-patel-9bab9639a
📝 Notes:
Greedy observation-based problem
Parity (odd/even) is the key concept
Moving by 2 positions is free
Simple and efficient one-pass solution
#leetcode #leetcode1217 #dsa #algorithms #coding #programming #codinginterview #interviewprep #greedy #math #leetcodeeasy #problemSolving #learncoding #codewithme #developerlife #tech #placements #softwareengineer #codinglife #100daysofcode
👍 Like | Comment | Subscribe for more LeetCode solutions!
Видео LeetCode 1217 | Minimum Cost to Move Chips to The Same Position | Easy Problem | C++ Solution канала Vishwa Patel
🚀 Problem: LeetCode 1217 – Minimum Cost to Move Chips to The Same Position
🧠 Approach:
Observe the movement cost rules:
Moving a chip by 2 positions costs 0
Moving a chip by 1 position costs 1
Count:
Number of chips at even positions
Number of chips at odd positions
To make all chips meet at the same position:
Move the smaller group to the larger group
Return the minimum of:
even count
odd count
⏱ Time Complexity: O(n)
📦 Space Complexity: O(1)
🔗 GitHub Repository:
https://github.com/VishwaPatel892/Leetcode_Solutions
👨💻 LeetCode Profile:
https://leetcode.com/u/vishwa0102/
💼 LinkedIn:
https://www.linkedin.com/in/vishwa-patel-9bab9639a
📝 Notes:
Greedy observation-based problem
Parity (odd/even) is the key concept
Moving by 2 positions is free
Simple and efficient one-pass solution
#leetcode #leetcode1217 #dsa #algorithms #coding #programming #codinginterview #interviewprep #greedy #math #leetcodeeasy #problemSolving #learncoding #codewithme #developerlife #tech #placements #softwareengineer #codinglife #100daysofcode
👍 Like | Comment | Subscribe for more LeetCode solutions!
Видео LeetCode 1217 | Minimum Cost to Move Chips to The Same Position | Easy Problem | C++ Solution канала Vishwa Patel
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13 ч. 30 мин. назад
00:04:06
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