# Min Moves To Obtain String Without 3 Identical Consecutive Letters

Min Moves To Obtain String Without 3 Identical Consecutive Letters – Given an array X[] of size n, we need to find the maximum and minimum elements contained in the array. Our algorithm should make the least number of comparisons.

Important Note: We recommend that students solve this problem before proceeding to the solutions. If it is solved, then good! We’d love to hear your ideas in the message below. Otherwise, no problem! Consider this a great opportunity to learn a new problem-solving model.

## Min Moves To Obtain String Without 3 Identical Consecutive Letters

Now we are moving forward and discussing solution ideas step by step. Following these steps will help you come up with an effective solution during a coding interview.

### Distance Measures In Data Science

In the worst case, we make two comparisons at each step of the iteration. This situation occurs when the array is sorted in descending order. In this case, the first if statement is false every time, and the second if statement is true every time. So the total number of comparisons in the worst case = 2*(n – 1) = 2n – 2

The best case occurs when the elements are sorted in ascending order. In this case, a total of n – 1 comparisons were made. Thinking.

Now the important question is: can we solve this problem in another way? can we think recursively to design an efficient solution? Here’s an idea!

To analyze the time complexity, let’s define a recurrence relation. Let T(n) be the problem size n time complexity.

#### Excel Get And Transform (power Query) Tutorial

We can solve this recurrence relation exactly using the recursion tree method. For better analysis, let n be a power of 2.

Note: Up to level (i – 1), each subproblem performs 2 comparisons in the merge step. The last level is the case of the base case, where only one comparison is made.

Note: Here the time complexity is also O(n), but the total number of comparison operation is less than the previous approach.

In the first approach, we perform two comparison operations for each element in the worst case. Now the important question — can we further optimize it and reduce the total number of comparison operations?

#### Multiple Choice Question

One idea :  Select elements in pairs and try to update min and max. Suppose the maximum and minimum up to the (i – 1)th index are updated in the max and min variables. Now we consider the i-th and (i + 1)-th index pair in the next iteration.

In both scenarios, we do 3 comparisons (in the worst case) to update the maximum and minimum 2 elements. In other words, we save one comparison with respect to the first approach, where 4 comparisons are required for 2 elements (in the worst case).

Initialization : If the array size is odd, we initialize the first element as min and max, and if the array size is even, we compare the first two elements and initialize min and max respectively.

For each pair, there are three comparisons: between the elements of the first pair, and the other two between min and max. Total number of comparisons = 3 * (n-1) / 2 (If n is odd) or 3n/2 – 2 (If n is even).

#### Numbers In Python

Time complexity = O(n). Here we notice that the total number of comparisons is smaller than in the first approach. In other words, pairwise comparison helps to further optimize the first approach. (Thinking)

Important Note: We recommend that students convert the above pseudocodes to their preferred programming language (C, C++, Java, Python, etc.) and test all test cases. Please let us know if you find any bugs or errors; We would be very grateful. Enjoy programming!

Space complexity = O(logn), Total number of comparisons = 3n/2 – 2 (If n is a power of 2)

## Using Python Datetime To Work With Dates And Times

In this technique, pointers represent the index or iteration attribute of a node, such as next. The steps in the two-point approach are:

Pointer initialization — Starting points. The metrics can be anywhere depending on what we are trying to achieve. On the left side of the image we have two pointers that start at the same position, i.e. the beginning of the linked list. In the right part of the picture, we have one of the pointers at the extreme ends at the initial index, and the other at the final index.

Indicator movement is what decides how we approach the solution. The pointer can move in one direction (left in the image above) or in the opposite direction (right in the image above). Also on the left side of the picture are different steps for the readings (top (slow) 1 unit bottom (fast) 2 units).

A stop condition is what determines when we stop. We continue until we reach a node whose next element is None in the left part. On the right we continue until the beginning is less than the end (i < j).

#### Sorting Algorithms Explained With Examples In Javascript, Python, Java, And C++

Given an integer array sorted in non-decreasing order, return the array of squares of each number sorted in non-decreasing order.

Given three sorted arrays A, B, and C, which do not necessarily have the same size. Compute the minimum absolute difference between the maximum and minimum number of any triplets A[i], B[j], C[k] such that they belong to the arrays A, B, and C, respectively, that is, minimize (max(A[i) ], B[j], C[k]) — min(A[i], B[j], C[k]))