LeetCode - Next Permutation
Problem statement
Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order).
The replacement must be in place and use only constant extra memory.
Problem statement taken from: https://leetcode.com/problems/next-permutation
Example 1:
Input: nums = [1, 2, 3]
Output: [1, 3, 2]
Example 2:
Input: nums = [3, 2, 1]
Output: [1, 2, 3]
Example 3:
Input: nums = [1, 1, 5]
Output: [1, 5, 1]
Example 4:
Input: nums = [1]
Output: [1]
Constraints:
- 1 <= nums.length <= 100
- 0 <= nums[i] <= 100
Explanation
Brute force approach
Brute force approach is to find all possible permutations of the array elements and find out the permutation which is the next largest one.
The problem here is, we are generating all permutations of the array elements and it takes lot of time.
The time complexity of this approach is O(N!) and space complexity is O(N).
Single pass approach
For a given sequence which is in descending order as below
[8, 5, 3, 2, 1]
there is no next larger permutation possible. This gives us a hint on identifying the next larger permutation.
We need to find the first pair of two successive numbers nums[i] and nums[i − 1], from the right, which satisfy nums[i] > nums[i − 1].
Once we find the index i - 1, we need to replace the number nums[i - 1] with the number which is just larger than itself among the numbers lying to its right section nums[i]..nums[nums.size() - 1], say nums[j].
We swap the numbers nums[i - 1] and nums[j]. We reverse all the numbers from index i and nums.size() - 1.
Algorithm
- return if nums.size() <= 1
- set n = nums.size(), i = n - 1
- loop while i > 0
- if nums[i] > nums[i - 1]
- break
- if i <= 0
- i = 0
- set x = ( i == 0 ) ? nums[i] : nums[i - 1]
- smallest = i
- loop for j = i + 1; j < n; j++
- nums[j] > x && nums[j] < nums[smallest]
- smallest = j
- swap(&nums[smallest], (i == 0 ? &nums[i] : &nums[i - 1]));
- sort(nums.begin() + i, nums.end());
C++ solution
class Solution {
public: void swap(int *a, int *b)
{
int temp = *a;
*a = *b;
*b = temp;
}
public:
void nextPermutation(vector<int>& nums) {
if(nums.size() <= 1){
return;
}
int n = nums.size();
int i = n - 1;
for(;i > 0; i--){
if(nums[i] > nums[i-1])
break;
}
if(i <= 0){
i = 0;
}
int x = (i == 0 ? nums[i] : nums[i - 1]);
int smallest = i;
for(int j = i + 1; j < n; j++){
if(nums[j] > x && nums[j] < nums[smallest])
smallest = j;
}
swap(&nums[smallest], (i == 0 ? &nums[i] : &nums[i - 1]));
// we can also use reverse
sort(nums.begin() + i, nums.end());
}
};Golang solution
func reverse(nums []int) {
for i := 0; i < len(nums); i++ {
j := len(nums) - 1 - i
if i >= j {
break
}
nums[i], nums[j] = nums[j], nums[i]
}
}
func nextPermutation(nums []int) {
i := 0
for i = len(nums) - 2; i >= 0; i-- {
if nums[i] < nums[i + 1] {
break
}
}
if i == -1 {
reverse(nums)
return
}
var j int
for j = len(nums)-1; j > i; j-- {
if nums[j] > nums[i] {
break
}
}
nums[i], nums[j] = nums[j], nums[i]
reverse(nums[i + 1:])
}Javascript solution
var nextPermutation = function(nums) {
if (nums === null || nums.length === 0) {
return nums;
}
let index = -1;
for (let i = nums.length - 2; i >= 0; i--) {
if (nums[i] < nums[i + 1]) {
index = i;
break;
}
}
if (index >= 0) {
for (let i = nums.length - 1; i > index; i--) {
if (nums[i] > nums[index]) {
let temp = nums[i];
nums[i] = nums[index];
nums[index] = temp;
break;
}
}
}
let start = index + 1;
let end = nums.length - 1;
while (start < end) {
let temp = nums[start];
nums[start] = nums[end];
nums[end] = temp;
start++;
end--;
}
};Dry Run
Let's dry-run our algorithm to see how the solution works.
Input: nums = [1, 2, 3, 6, 5, 4]
Output: [1, 2, 4, 3, 5, 6]
Step 1: nums.size() <= 1
6 <= 1
false
Step 2: n = nums.size()
n = 6
i = n - 1
= 6 - 1
= 5
Step 3: loop for i > 0
5 > 0
true
if nums[i] > nums[i - 1]
nums[5] > nums[4]
4 > 5
false
i--
i = 4
Step 4: loop for i > 0
4 > 0
true
if nums[i] > nums[i - 1]
nums[4] > nums[3]
5 > 6
false
i--
i = 3
Step 5: loop for i > 0
3 > 0
true
if nums[i] > nums[i - 1]
nums[3] > nums[2]
6 > 3
true
break
Step 6: i <= 0
3 <= 0
false
Step 7: x = (i == 0 ? nums[i] : nums[i - 1])
= (3 == 0 ? nums[3] : nums[2])
= (false ? nums[3] : nums[2])
= nums[2]
= 3
smallest = i
= 3
Step 8: loop for(j = i + 1; j < n; j++)
j = 3 + 1
= 4
j < n
4 < 6
true
nums[j] > x && nums[j] < nums[smallest]
nums[4] > 3 && nums[4] < nums[3]
5 > 3 && 5 < 6
true
smallest = j
= 4
j++
j = 5
Step 9: loop for(j = i + 1; j < n; j++)
j < n
5 < 6
true
nums[j] > x && nums[j] < nums[smallest]
nums[5] > 3 && nums[5] < nums[4]
4 > 3 && 4 < 6
true
smallest = j
= 5
j++
j = 6
Step 10: loop for(j = i + 1; j < n; j++)
j < 6
6 < 6
false
Step 11: swap(&nums[smallest], (i == 0 ? &nums[i] : &nums[i - 1]));
swap(&nums[5], 3 == 0 ? &nums[3] : &nums[2])
swap(&nums[5], &nums[2])
swap(3, 4)
[1, 2, 4, 6, 5, 3]
Step 12: reverse(nums[i], nums[n - 1])
reverse(nums[3], nums[5])
[1, 2, 4, 3, 5, 6]