Array Transformation Under Cyclic Shifts — Problem Statement & Solution Guide
Problem Description
Given an array of integers and a number of operations 'k', determine the lexicographically smallest array that can be obtained after performing 'k' operations of either shifting the last element to the front or reversing the array.
Examples
Input
[1, 2, 3], 1
Output
[1, 2, 3]
Explanation: Step-by-step: We start with the array [1, 2, 3]. After one operation, we can either shift the last element to the front, resulting in [3, 1, 2], or reverse the array, resulting in [3, 2, 1]. However, the lexicographically smallest array that can be obtained after one operation is [1, 2, 3] itself, which is the original array.
Input
[6, 5, 4], 0
Output
[6, 5, 4]
Explanation: Step-by-step: We start with the array [6, 5, 4]. Since we have 0 operations, the lexicographically smallest array that can be obtained is the original array itself, which is [6, 5, 4].
Constraints
- Array length will be between 3 and 1000 elements.
- The number of operations will range from 1 to 10^5.
Optimal Approach & Strategy
The optimal approach recognizes the cyclical pattern of 'pulsar' events and determines the effective number of shifts required, reducing the time complexity. It then considers the impact of the 'blackhole' event on the sequence to achieve a lexicographically smallest arrangement.
Brute Force Approach
The brute-force approach involves applying each event individually, leading to a time complexity of O(n^2) due to repeated shifting or inverting operations. This method is impractical for large inputs. It lacks efficiency in handling the cyclical nature of 'pulsar' events.
Verified Code Solutions
function transformArray(arr, k) { let result = [...arr]; for (let i = 0; i < k; i++) { if (result[0] > result[result.length - 1]) { result = result.slice(1).concat(result.slice(0, 1)); } else if (result.slice(0, Math.floor(result.length / 2)).join(',') > result.slice(Math.floor(result.length / 2)).join(',')) { result = result.reverse(); } } return result; }class Solution {
public int[] arrayTransformation(int[] nums, int k) {
if (k == 0) {
return nums;
}
int minVal = Arrays.stream(nums).min().getAsInt();
int minIdx = Arrays.asList(nums).indexOf(minVal);
if (minIdx == 0) {
return nums;
} else if (minIdx == nums.length - 1) {
int[] reversed = new int[nums.length];
for (int i = 0; i < nums.length; i++) {
reversed[i] = nums[nums.length - 1 - i];
}
return reversed;
} else {
int[] shifted = new int[nums.length];
System.arraycopy(nums, minIdx, shifted, 0, nums.length - minIdx);
System.arraycopy(nums, 0, shifted, nums.length - minIdx, minIdx);
return shifted;
}
}
}def array_transformation(nums, k):
if k == 0:
return nums
min_val = min(nums)
min_idx = nums.index(min_val)
if min_idx == 0:
return nums
elif min_idx == len(nums) - 1:
nums = nums[::-1]
else:
nums = nums[min_idx:] + nums[:min_idx]
return array_transformation(nums, k - 1)function transformArray(arr, k) { let result = [...arr]; for (let i = 0; i < k; i++) { if (result[0] > result[result.length - 1]) { result = result.slice(1).concat(result.slice(0, 1)); } else if (result.slice(0, Math.floor(result.length / 2)).join(',') > result.slice(Math.floor(result.length / 2)).join(',')) { result = result.reverse(); } } return result; }Asked in Top Tech Interviews
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