I need to know minimum moves not swaps to make a equal to b. With swapping approach, the minimum swaps is 2, but I need minimum moves which in this case is 1; by moving 'A' to the front of list.
For the elements not in LCS, you can correct its position by just one move. Learn more. Minimum moves not swaps to make two arrays equal Ask Question. Asked 1 year, 3 months ago.
Active 1 year, 3 months ago. Viewed times. It sounds this topic is a duplicate but it's not. Mahdi Mahdi 61 7 7 bronze badges. What have you tried so far? DanielMesejo I don't mean replacing. Yes, You are right! Kaidul yes, you can move any element to anywhere! Active Oldest Votes. Index Ama Index Ama 69 2 2 bronze badges. Sign up or log in Sign up using Google.
Problem Statement: We have two strings A and B with the same super set of characters. We need to change these strings to obtain two equal strings.
Leetcode 453. Minimum Moves to Equal Array Elements
In each move we can perform one of the following operations:. A move can be performed on either string. What is the minimum number of moves that we need in order to obtain two equal strings? Input Format and Constraints: The first and the second line of the input contains two strings A and B. It is guaranteed that the superset their characters are equal.
Output Format: Print the minimum number of moves to the only line of the output. Explanation: Swap the first and last character of the string aab to convert it to baa. The two strings are now equal. Can someone guide me what is wrong in my approach.
The proper algorithm is probably even more complex, but you can see now what's wrong in your solution. The initial node will be the original string.
The goal node will be the target string. Each child of a node will be all possible transformations of that string. The current cost g x is simply the number of transformations thus far.
The heuristic h x is half the number of characters in the wrong position.Given a non-empty integer array of size nfind the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1. Thanks for using LeetCode!
To view this solution you must subscribe to premium. When a star is present, we may need to check many different suffixes of the text and see if they match the rest of the pattern. A recursive solution is a straightforward way to represent this relationship. Then, we may ignore this part of the pattern, or delete a matching character in the text. If we have a match on the remaining strings after any of these operations, then the initial inputs matched.
Space Complexity: For every call to matchwe will create those strings as described above, possibly creating duplicates. As the problem has an optimal substructureit is natural to cache intermediate results. We can describe our answer in terms of answers to questions involving smaller strings.
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Approach 2: Dynamic Programming Intuition As the problem has an optimal substructureit is natural to cache intermediate results. Sign in to view your submissions. Sign in. Console Contribute.
Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1.
I tried to brute force it, but I couldn't come with a correct algorithm, the loop invariants are incorrect. Would someone fix it with explanation so that I can improve my algorithms skills? I will solve it as below:. Firstly, you don't need to do brute force for this problem.
There is a linear time solution and the answer is. Incrementing all but one is equivalent to decrementing that one. So let's do that instead. How many single-element decrements to make all equal?
No point to decrementing below the current minimum, so how many single-element decrements to make all equal to the current minimum? Just take the difference from what's currently there the sum to what we want n times the minimum. Learn more.
Asked 3 years, 5 months ago. Active 2 years, 1 month ago. Viewed 4k times. Active Oldest Votes.From a non-empty integer array of length n, find the minimum number of steps that can make the values of each element in the array equal. One step is to select an n-1 element in the array plus one.
For example, the minimum number of steps required to achieve an equal number of [1, 2, 3] is three steps. The process is as follows:. Assuming that the minimum value of an array with n elements is min, the sum of all elements of the array is sum, the minimum step to achieve equality is move, and the average value is target, the formula can be obtained.
It can be concluded by calculating both of them. If not every step will add the minimum value to one, then this value will always be the minimum value, and it will never reach the final target value. Conversely, the title is equivalent to starting from the target value, each step will be a value — 1, until returning to the initial array, then each time is executed — 1, the result is the minimum value of the array.Frequently Asked Java Program 16: How To Check The Equality Of Two Arrays
Leetcode Tags: javaleetcodemath. Objective-C design pattern analysis strategy. Longest Palindromic Subsequence.
Making elements of two arrays same with minimum increment/decrement
Pre: Advanced features, modules and IO operations of Python. Next: Ajax implements the effect of input prompt.Given two arrays which have same values but in different order, we need to make second array same as first array using minimum number of swaps. This problem can be solved by modifying the array B. We save the index of array A elements in array B i.
This modified array represents distribution of array A element in array B and our goal is to sort this modified array in minimum number of swaps because after sorting only array B element will be aligned with array A elements.
Now count of minimum swaps for sorting an array can be found by visualizing the problem as a graphthis problem is already explained in previous article. So we count these swaps in modified array and that will be our final answer. Please see below code for better understanding. This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.
Code Review Stack Exchange is a question and answer site for peer programmer code reviews. It only takes a minute to sign up. Long story short, the other day I discovered the site LeetCode and started solving some of its problems until I stumbled onto this one.
Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1. The problem is, the solution I am trying to submit always exceeds the allowed processing time limit and I feel like I've reached a dead end. Can anyone suggest a way to improve my code? Do you see why these are the same? Suppose we start with 1, 2, 3. Adding 1, 1, 0 produces 2, 3, 3.
Adding 0, 0, -1 produces 1, 2, 2which is effectively the same thing. However many moves it takes to make 2, 3, 3 all equal is surely the same number of moves as it takes to make 1, 2, 2all equal, or 0, 1, 1or,or whatever. Where you start is irrelevant. Once you know that the problem is actually "how many single decrements does it take to get all the elements equal?
There is never any point in decrementing the smallest element, and every element will have to eventually equal the smallest element. Therefore the answer is: it's the sum of the differences of every element from the minimal element. Though as noted in the comments, this second version can overflow. Consider doing all the math in longs or BigIntegers if you're worried about overflow cases. Firstly, identify hidden loops, such as MinMax and IndexOfand merge those together - as per Paparazzi's answer.
There's no need to count to n 3 times in a row. The question asks you to find the number of moves, not to actually execute the number of moves.
You've already optimized this by working with diff rather than adding just 1 each time. However, this is an optimization for the best-case scenario e. Further optimization could be achieved by removing the lowest duplicated numbers - though this once again is a best-case optimization. Finally, rethink the whole thing.