Next Greater Element III LeetCode Solution

Last updated on July 18th, 2024 at 04:51 am

Here, We see Next Greater Element III LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

String

Companies

Bloomberg

Level of Question

Medium

Next Greater Element III LeetCode Solution

Next Greater Element III LeetCode Solution

Problem Statement

Given a positive integer n, find the smallest integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive integer exists, return -1.

Note that the returned integer should fit in 32-bit integer, if there is a valid answer but it does not fit in 32-bit integer, return -1.

Example 1:
Input: n = 12
Output: 21

Example 2:
Input: n = 21
Output: -1

1. Next Greater Element III LeetCode Solution C++

class Solution {
public:
    int nextGreaterElement(int n) {
        vector<int> digits;
        while (n) {
            digits.push_back(n % 10);
            n /= 10;
        }
        for (int i = 1, j, len = digits.size(), lmt = INT_MAX / 10; i < len; i++) {
            if (digits[i - 1] > digits[i]) {
                j = 0;
                while (digits[j] <= digits[i]) j++;
                swap(digits[j], digits[i]);
                sort(begin(digits), begin(digits) + i, greater<int>());
                while (digits.size()) {
                    n = n * 10 + digits.back();
                    digits.pop_back();
                    if (n >= lmt && digits.size()) return -1;
                }
                return n;
            }
        }
        return -1;        
    }
};

2. Next Greater Element III LeetCode Solution Java

class Solution {
    public int nextGreaterElement(int n) {
        char arr[] = (Integer.toString(n)).toCharArray();
        int i=arr.length-2;
        StringBuilder ans = new StringBuilder();
        while(i>=0 && arr[i] >= arr[i+1])
            i--;
        if(i == -1)
            return -1;
        int k = arr.length-1;
        while(arr[k] <= arr[i])
            k--;
        swap(arr,i,k);
        for(int j=0;j<=i;j++)
            ans.append(arr[j]);
        for(int j=arr.length-1;j>i;j--)
            ans.append(arr[j]);
        long ans_ = Long.parseLong(ans.toString());
        return (ans_ > Integer.MAX_VALUE) ? -1 : (int)ans_;
    }
    void swap(char[] arr,int i,int j)
    {
        char temp = arr[j];
        arr[j] = arr[i];
        arr[i] = temp;
    }
}

3. Next Greater Element III Solution JavaScript

var nextGreaterElement = function(n) {
    var numbers = ("" + n).split('').map(n=>parseInt(n));
    var lastIncreasePosition = null;
    var smallestToSwapWithPosition;
    var i, l;
    for (i = 0, l = numbers.length; i < l; i++) {
        if (numbers[i] < numbers[i + 1] && i + 1 < l) {
            lastIncreasePosition = i;
            smallestToSwapWithPosition = i + 1;
        } else if (lastIncreasePosition !== null) {
            if (numbers[smallestToSwapWithPosition] > numbers[i] && numbers[i] > numbers[lastIncreasePosition]) {
                smallestToSwapWithPosition = i;
            }
        }
    }
    if (lastIncreasePosition === null) {
        return -1;
    }
    let temp = numbers[lastIncreasePosition];
    numbers[lastIncreasePosition] = numbers[smallestToSwapWithPosition];
    numbers[smallestToSwapWithPosition] = temp
    var sorted = numbers.slice(0, lastIncreasePosition + 1)
    var unsorted = numbers.slice(lastIncreasePosition + 1);
    unsorted = unsorted.sort((a,b) => a - b);
    numbers = [...sorted, ...unsorted];
    n = parseInt(numbers.join(''));
    if (n > 2147483647) {
        return -1;
    }
    return n;
};

4. Next Greater Element III Solution Python

class Solution(object):
    def nextGreaterElement(self, n):
        nums = list(str(n))
        i = len(nums) - 1
        while i > 0 and nums[i-1] >= nums[i]:
            i -= 1
        if i == 0:
            return -1
        j = len(nums) - 1
        while j > i-1 and nums[j] <= nums[i-1]:
            j -= 1
        nums[i-1], nums[j] = nums[j], nums[i-1]
        nums[i:] = nums[i:][::-1]
        res = int(''.join(nums))
        return res if res < 2**31 else -1
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