Next Permutation Problem in Kotlin

The next permutation problem is a classic problem in computer science that involves finding the lexicographically next permutation of a given sequence of numbers. In other words, given a sequence of numbers, the goal is to find the smallest possible sequence that comes after the original sequence in lexicographic order.

For example, if the original sequence is [1, 2, 3], the next lexicographically greater permutation is [1, 3, 2]. If there is no such permutation (i.e., the original sequence is already the largest possible permutation), then the algorithm should return the first permutation in lexicographic order.

The next permutation problem is commonly used in a variety of algorithms, such as the permutation-based combinatorial optimization problems and the sorting algorithms that use comparison-based sorting methods.

The algorithm to find the next permutation involves the following steps:

1. Starting from the right end of the sequence, find the first element that is smaller than the element to its right. If no such element exists, the sequence is already in its largest permutation, and we return the first permutation in lexicographic order.
2. Starting from the right end of the sequence, find the smallest element that is greater than the element found in step 1.
3. Swap the elements found in steps 1 and 2.
4. Reverse the sequence starting from the element immediately to the right of the element found in step 1.

Here is an example implementation of the algorithm in Kotlin:

fun nextPermutation(nums: IntArray) {
    var i = nums.size - 2
    while (i >= 0 && nums[i] >= nums[i + 1]) {
        i--
    }
    if (i >= 0) {
        var j = nums.size - 1
        while (j >= 0 && nums[j] <= nums[i]) {
            j--
        }
        swap(nums, i, j)
    }
    reverse(nums, i + 1)
}
fun swap(nums: IntArray, i: Int, j: Int) {
    val temp = nums[i]
    nums[i] = nums[j]
    nums[j] = temp
}
fun reverse(nums: IntArray, start: Int) {
    var i = start
    var j = nums.size - 1
    while (i < j) {
        swap(nums, i, j)
        i++
        j--
    }
}
fun main() {
    val nums = intArrayOf(1, 2, 3)
    nextPermutation(nums)
    println(nums.contentToString()) // prints "[1, 3, 2]"

    val nums2 = intArrayOf(3, 2, 1)
    nextPermutation(nums2)
    println(nums2.contentToString()) // prints "[1, 2, 3]"

    val nums3 = intArrayOf(1, 1, 5)
    nextPermutation(nums3)
    println(nums3.contentToString()) // prints "[1, 5, 1]"
}

Here’s a detailed explanation of the steps in the nextPermutation function:

1. We start by initializing a variable i to the second-to-last index in the input array. We then iterate backwards through the array from right to left until we find the first index i where the value at i is less than the value at the next index i + 1, or until we reach the beginning of the array. At the end of this step, we know that the subarray nums[i+1:] is sorted in decreasing order.
2. If we found an index i in step 1, we then initialize another variable j to the last index in the array, and iterate backwards through the array again until we find the first index j where the value at j is greater than the value at i. At the end of this step, we know that the value at j is the smallest value in the subarray nums[i+1:] that is greater than the value at i.
3. We swap the values at indices i and j in the array.
4. Finally, we reverse the subarray nums[i+1:]. This step is necessary because we know that nums[i+1:] is sorted in decreasing order, and we want to find the smallest possible permutation that comes after the current permutation in lexicographic order. By reversing the subarray, we can ensure that the resulting permutation is as small as possible.

The time complexity of the nextPermutation function is O(n), where n is the length of the input array. This is because the function performs a constant number of operations for each element in the array. In particular, it iterates through the array twice in the worst case (once to find the first index i, and once to find the first index j), and then performs a constant number of operations to swap the values at indices i and j and reverse the subarray nums[i+1:].

The space complexity of the nextPermutation function is O(1), because the function does not create any additional data structures or use any significant amount of extra memory. The function performs all of its operations in place on the input array nums. Therefore, the amount of memory used by the function is independent of the size of the input array, and is bounded by a constant amount of memory regardless of the size of the input.