Next Integer Permutation - Amazon Top Interview Questions

Problem Statement :

```Given an integer n, return the next bigger permutation of its digits.

If n is already in its biggest permutation, rotate to the smallest permutation.

Constraints

n < 2 ** 31

Example 1

Input

n = 527

Output

572

Example 2

Input

n = 321

Output

123

Explanation

321 is already the biggest permutation so it rotates to the smallest.

Example 3

Input

n = 20

Output

2```

Solution :

```                        ```Solution in C++ :

int solve(int n) {
string s = to_string(n);
int i = 0, j = 0;
for (i = s.size() - 2; i >= 0; i--) {
if (s[i] < s[i + 1]) break;
}
if (i < 0) {
sort(s.begin(), s.end());
return stoi(s);
}
for (j = s.size() - 1; j >= 0; j--) {
if (s[j] > s[i]) break;
}
swap(s[i], s[j]);
sort(s.begin() + i + 1, s.end());
return stoi(s);
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int n) {
ArrayList<Integer> ar = new ArrayList<>();
int a = n;
while (a > 0) {
a = a / 10;
}
int arr[] = new int[ar.size()];
int j = ar.size() - 1;
for (int x : ar) {
arr[j--] = x;
}
int mark = arr.length - 2;
while (mark >= 0 && arr[mark + 1] <= arr[mark]) {
mark--;
}
if (mark == -1) {
reverse(arr, 0);
return number(arr);
}
if (mark >= 0) {
int k = arr.length - 1;
while (k >= mark && arr[k] <= arr[mark]) {
k--;
}
swap(arr, mark, k);
}
reverse(arr, mark + 1);
return number(arr);
}

private int number(int nums[]) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
sum = sum * 10 + nums[i];
}
return sum;
}
private void reverse(int nums[], int start) {
int i = start, j = nums.length - 1;
while (i < j) {
swap(nums, i, j);
i++;
j--;
}
}

private void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, num):
# convert the number to a list of each digit
A = list(map(int, str(num)))

i = len(A) - 1

# find the position of the rightmost element where we encounter
# increasing comparison (A[i-1] < A[i])
# for 527, this will be 27 and i = 2, A[i] = 7
while i > 0 and A[i - 1] >= A[i]:
i -= 1

if i == 0:
# if i == 0, it means the whole list was sorted in reverse order
# in other words, it's already the highest permutation
# so we just reverse it to get the smallest permutation
A.reverse()
else:
j = len(A) - 1
k = i - 1  # for 527: k = 1, A[k] = 2

# starting from the position i-1, we go left to find the number
# that's larger than the that number
while A[j] <= A[k]:
j -= 1
A[j], A[k] = A[k], A[j]  # for 527, we swap 2 and 7 since j=2

# we reverse the remaining part of the array
# after the index k + 1, since it should be reset to
# the smallest permutation
left, right = k + 1, len(A) - 1
while left < right:
A[left], A[right] = A[right], A[left]
left, right = left + 1, right - 1

return int("".join(map(str, A)))```
```

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