Intervals Between Identical Elements - Leetcode Solution

The problem statement for Intervals Between Identical Elements on LeetCode is as follows:

You are given a 0-indexed integer array nums of length n containing integers from 1 to n. In one operation, you can select an element of the array and increment it by 1.

For example, if nums = [1,2,3], you can:

• Increment nums[0], resulting in nums = [2,2,3].
• Increment nums[1], resulting in nums = [1,3,3].
• Increment nums[2], resulting in nums = [1,2,4].

All elements of nums are distinct.

Return the minimum number of operations to move every element to the same value. Elements may only be incremented.

Example 1:

Input: nums = [1,2,2] Output: 1 Explanation: Incrementing nums[0] gives nums = [2,2,2].

Example 2:

Input: nums = [2,3,3,2,4] Output: 2 Explanation: Incrementing nums[0] and nums[3] gives nums = [3,3,3,3,4].

Approach:

The basic idea to solve the problem is to find all the intervals of identical elements and calculate the number of operations required to move all elements to the same value. The minimum number of operations required will be the sum of operations required for each interval.

For example, if we have an array nums = [1,2,2,3,4,4,4], the intervals of identical elements would be:

• Interval 1: [2,2] with 1 operation required to move all elements to 2
• Interval 2: [4,4,4] with 2 operations required to move all elements to 4

Therefore, the minimum number of operations required would be 1 + 2 = 3.

Algorithm:

• Initialize a variable count to 0.
• Initialize two variables start and end to -1.
• Iterate through the array nums from left to right.
• If the current element is equal to the next element, update the end index to the current index.
• If the current element is not equal to the next element, calculate the operations required to move all elements in the interval [start+1, end] to the same value and add it to count.
• Update start and end to -1.
• Return count.

Code:

Here is the Python code to solve the problem:

class Solution: def minOperations(self, nums: List[int]) -> int: count = 0 start, end = -1, -1

    for i in range(len(nums)-1):
        if nums[i] == nums[i+1]:
            end = i+1
        else:
            if start != -1 and end != -1:
                count += (end - start)
            start, end = -1, -1
    
    if start != -1 and end != -1:
        count += (end - start)
    
    return count

1