Given a positive 32-bit integer n, you need to find the smallest 32-bit integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive 32-bit integer exists, you need to return -1.
Example 1:
Input: 12
Output: 21Example 2:
Input: 21
Output: -1
Approach #1: String. [Java]
class Solution {
public int nextGreaterElement(int n) {
int i, j;
char[] nums = new String(n + "").toCharArray();
for (i = nums.length-1; i > 0; --i) {
if (nums[i-1] < nums[i])
break;
} if (i == 0) return -1; int x = nums[i-1], smallest = i;
for (j = i + 1; j < nums.length; ++j) {
if (nums[j] > x && nums[j] < nums[smallest])
smallest = j;
}
char temp = nums[i-1];
nums[i-1] = nums[smallest];
nums[smallest] = temp;
Arrays.sort(nums, i, nums.length);
long ret = Long.parseLong(new String(nums)); return ret < Integer.MAX_VALUE ? (int)ret : -1;
}
}
Analysis:
At first, let’s look at the edge cases:
1. If all digits sorted in descending order, then output is always “Not Possible”. Foe example 4321.
2. If all digits are sorted in ascending order, then we need to swap last two digits. For example, 1234.
3. For other cases, we need to process the number from rightmost side.
The main algorithm works in following steps:
1. Traverse the given number from rightmost digit, keep traversing till you find a digit which is smaller the given trversed digit. For example, if the input number is “534976”, we stop at 4 because 4 is smaller than next digit 9. If we do not find such a digit, then outpit is “Not Possible”.
2. Now search the right side of above found digit ‘d’ for the smallest digit greater than ‘d’. For “534976”, the right side of 4 contains “976”. The smallest digit greater than 4 to 6.
3. Swap the above found two digits, we get 536974 in above example.
4. Now sort all digits form position next to ‘d’ to the end of number. The number that we get after sorting is the output. For above example, we sort digits in bold 53974. We get “536479” which is the next greater number for input 534976.
Reference:
