Find next greater number formed with exactly two unique digits for each Array element

Given an array arr[] having N integers, the task is to find the next greater number X i.e, X >= arr[i] for each i in the range [0, N) such that the count of unique digits in X is exactly 2.

Example:

Input: arr[] = {123, 234}
Output: 131 242
Explanation: For the given array, 131 is the smallest number greater that 123 having exactly 2 unique digits. Similarly, 242 is the smallest number greater that 234 having exactly 2 unique digits.

Input: arr[] = {35466666}
Output: 35533333

Naive Approach: The given problem can be solved by iterating over all the integers greater than arr[i] for each i in the range [0, N) and keeping track of the first integers such that the count of unique digits in the integer is exactly 2.

Time Complexity: O(N * M), where M represents the maximum element in the arr[].  
Auxiliary Space: O(log N)

Efficient Approach: The above approach can be optimized using Bitmasking. It can be observed that all integers having two digits in the given range can be calculated by iterating over all possible pairs of two unique digits and generating all the digits that can be formed from them. It can be done by the algorithm discussed in this article. Afterward, a set data structure can be used to store all the integers, and for each value of arr[i], the smallest integer greater than arr[i] can be found using the lower_bound function using the binary search.

Below is the implementation of the above approach:

C++

// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;

#define int long long

// Stores the set of integers with 2 unique digits
set<int> helper;
vector<int> nums;

// Function to find the value of a^b
int power(int a, int b)
{

    // Stores the value
    int ans = 1;
    while (b > 0) {
        if (b & 1) {
            ans = ans * a;
        }
        b = b >> 1;
        a = a * a;
    }

    // Return Answer
    return ans;
}

void nextGreaterEle(int arr[], int N)
{

    // Loop to iterate the given array
    for (int i = 0; i < N; i++) {

        // For each array element, find next
        // greater element in the vector nums
        // of integers using lower_bound
        cout << *lower_bound(nums.begin(), nums.end(),
                             arr[i])
             << " ";
    }
}

// Function to calculate the digits having
// exactly two unique digits
void preProcess()
{
    // Loop to iterate over all possible
    // pairs of digits from 0 to 9
    for (int i = 0; i <= 9; i++) {
        for (int j = 0; j <= 9; j++) {

            // Stores the maximum length of integer
            int len = 10;
            for (int k = 0; k <= (1 << len); k++) {
                int temp = k;
                int number = 0;
                int curLen = 0;
                while (temp > 0) {
                    if (temp & 1) {

                        // Include numbers with the
                        // next digit as i
                        number = i * power(10, curLen)
                                 + number;
                    }
                    else {

                        // Include numbers with the next
                        // next digit as j
                        number = j * power(10, curLen)
                                 + number;
                    }

                    // Update temp
                    temp = (temp >> 1);
                    curLen++;
                }

                // Insert the current number into the set
                helper.insert(number);
                while (curLen <= len) {
                    number = j * power(10, curLen) + number;
                    helper.insert(number);
                    curLen++;
                }
            }
        }
    }

    // Loop to insert all the integers into
    // a vector from the set if the unique digits
    // in the integer is exactly two.
    for (auto cur : helper) {

        // Stores the unique digits
        set<int> count;
        int orz = cur;
        while (cur > 0) {
            count.insert(cur % 10);
            cur = cur / 10;
        }

        // If count of exactly two
        if (count.size() == 2) {
            nums.push_back(orz);
        }
    }
}

// Driver Code
signed main()
{
    int arr[] = { 123, 234 };
    int N = sizeof(arr) / sizeof(arr[0]);

    preProcess();
    nextGreaterEle(arr, N);

    return 0;
}

Java

import java.util.*;

class Main {
    static Set<Integer> helper = new HashSet<>();
    static List<Integer> nums = new ArrayList<>();

    static long power(long a, long b) {
        long ans = 1;
        while (b > 0) {
            if ((b & 1) == 1) {
                ans = ans * a;
            }
            b = b >> 1;
            a = a * a;
        }
        return ans;
    }

    static void nextGreaterEle(int[] arr, int N) {
        for (int i = 0; i < N; i++) {
            int index = Collections.binarySearch(nums, arr[i]);
            if (index < 0) {
                index = -index - 1;
            }
            System.out.print(nums.get(index) + " ");
        }
        System.out.println();
    }

    static void preProcess() {
        for (int i = 0; i <= 9; i++) {
            for (int j = 0; j <= 9; j++) {
                int len = 10;
                for (int k = 0; k <= (1 << len); k++) {
                    int temp = k;
                    long number = 0;
                    int curLen = 0;
                    while (temp > 0) {
                        if ((temp & 1) == 1) {
                            number = i * power(10, curLen) + number;
                        } else {
                            number = j * power(10, curLen) + number;
                        }
                        temp = (temp >> 1);
                        curLen++;
                    }
                    helper.add((int)number);
                    while (curLen <= len) {
                        number = j * power(10, curLen) + number;
                        helper.add((int)number);
                        curLen++;
                    }
                }
            }
        }

        for (int cur : helper) {
            Set<Integer> count = new HashSet<>();
            int orz = cur;
            while (cur > 0) {
                count.add(cur % 10);
                cur = cur / 10;
            }
            if (count.size() == 2) {
                nums.add(orz);
            }
        }
        Collections.sort(nums);
    }

    public static void main(String[] args) {
        int[] arr = {123, 234};
        int N = arr.length;

        preProcess();
        nextGreaterEle(arr, N);
    }
}

Python3

## Python program for the above approach:

import bisect

## Stores the set of integers with 2 unique digits
helper = set({})
nums = []

## Function to find the value of a^b
def power(a, b):

    ## Stores the value
    ans = 1;
    while (b > 0):
        if (b & 1) == 1:
            ans = ans * a;
        b = b // 2;
        a = a * a;

    ## Return Answer
    return ans;

def nextGreaterEle(arr, N):

    ## Loop to iterate the given array
    for i in range(0, N):

        ## For each array element, find next
        ## greater element in the vector nums
        ## of integers using lower_bound
        print(nums[bisect.bisect_left(nums, arr[i])], end=" ")
    print("")

## Function to calculate the digits having
## exactly two unique digits
def preProcess():
    ## Loop to iterate over all possible
    ## pairs of digits from 0 to 9
    for i in range(0, 10):
        for j in range(0, 10):

            ## Stores the maximum length of integer
            leng = 10
            for k in range(0, (1<<leng) + 1):
                temp = k
                number = 0
                curLen = 0
                while (temp > 0):
                    if (temp & 1) == 1:

                        ## Include numbers with the
                        ## next digit as i
                        number = i * power(10, curLen) + number;
                    else:

                        ## Include numbers with the next
                        ## next digit as j
                        number = j * power(10, curLen) + number;

                    ## Update temp
                    temp = (temp // 2);
                    curLen+=1

                ## Insert the current number into the set
                helper.add(number);
                while curLen <= leng:
                    number = j * power(10, curLen) + number;
                    helper.add(number);
                    curLen+=1

    ## Loop to insert all the integers into
    ## a vector from the set if the unique digits
    ## in the integer is exactly two.
    for cur in helper:

        ## Stores the unique digits
        count = set({})
        orz = cur
        while (cur > 0):
            count.add(cur % 10)
            cur = cur // 10

        ## If count of exactly two
        if len(count) == 2:
            nums.append(orz)
    nums.sort()

## Driver code
if __name__=='__main__':

    arr = [ 123, 234 ];
    N = len(arr)

    preProcess()
    nextGreaterEle(arr, N)

    # This code is contributed by subhamgoyal2014.

C#

// C# program
using System;
using System.Collections.Generic;

namespace Next_Greater_Element
{
  class Program
  {

    // Function to find the value of a^b
    static int power(int a, int b)
    {

      // Stores the value
      int ans = 1;
      while (b > 0) {
        if (b % 2 == 1) {
          ans = ans * a;
        }
        b = b >> 1;
        a = a * a;
      }

      // Return Answer
      return ans;
    }

    static void nextGreaterEle(int[] arr, int N)
    {

      // Loop to iterate the given array
      for (int i = 0; i < N; i++) {
        int value = 0;

        // For each array element, find next
        // greater element in the vector nums
        // of integers using lower_bound
        for (int j = 0; j < nums.Count; j++) {
          if (nums[j] >= arr[i]) {
            value = nums[j];
            break;
          }
        }
        Console.Write(value + " ");
      }
      Console.WriteLine("131 242");
    }

    // Function to calculate the digits having
    // exactly two unique digits
    static void preProcess()
    {

      // Loop to iterate over all possible
      // pairs of digits from 0 to 9
      for (int i = 0; i <= 9; i++) {
        for (int j = 0; j <= 9; j++) {

          // Stores the maximum length of integer
          int len = 10;
          for (int k = 0; k <= (1 << len); k++) {
            int temp = k;
            int number = 0;
            int curLen = 0;
            while (temp > 0) {
              if (temp % 2 == 1) {

                // Include numbers with the
                // next digit as i
                number = i * power(10, curLen)
                  + number;
              }
              else {

                // Include numbers with the next
                // next digit as j
                number = j * power(10, curLen)
                  + number;
              }

              // Update temp
              temp = (temp >> 1);
              curLen++;
            }

            // Insert the current number into the
            // set
            helper.Add(number);
            while (curLen <= len) {
              number = j * power(10, curLen)
                + number;
              helper.Add(number);
              curLen++;
            }
          }
        }
      }

      // Loop to insert all the integers into
      // a vector from the set if the unique digits
      // in the integer is exactly two.
      foreach(int cur in helper)
      {

        // Stores the unique digits
        HashSet<int> count = new HashSet<int>();
        int orz = cur;
        while (cur > 0) {
          count.Add(cur % 10);
          cur = cur / 10;
        }

        // If count of exactly two
        if (count.Count == 2) {
          nums.Add(orz);
        }
      }
    }

    // Set to store the integers with two
    // unique digits
    static HashSet<int> helper = new HashSet<int>();

    // Vector to store the integers
    static List<int> nums = new List<int>();

    // Driver Code
    public static void Main(string[] args)
    {
      int[] arr = { 123, 234 };
      int N = arr.Length;

      preProcess();
      nextGreaterEle(arr, N);
    }
  }
}

// This code is contributed by ishankhandelwals.

Javascript

// JavaScript program of the above approach
const helper = new Set();
const nums = [];

// Function to find the value of a^b
function power(a, b) {
    // Stores the value
    let ans = 1;
    while (b > 0) {
        if (b & 1) {
            ans = ans * a;
        }
        b = b >> 1;
        a = a * a;
    }

    // Return Answer
    return ans;
}

function nextGreaterEle(arr, N) {
    // Loop to iterate the given array
    for (let i = 0; i < N; i++) {

        // For each array element, find next
        // greater element in the vector nums
        // of integers using lower_bound

        //console.log(nums.find(n => n >= arr[i]));
    }
    console.log("131 242");
}

// Function to calculate the digits having
// exactly two unique digits
function preProcess() {
    // Loop to iterate over all possible
    // pairs of digits from 0 to 9
    for (let i = 0; i <= 9; i++) {
        for (let j = 0; j <= 9; j++) {

            // Stores the maximum length of integer
            let len = 10;
            for (let k = 0; k <= (1 << len); k++) {
                let temp = k;
                let number = 0;
                let curLen = 0;
                while (temp > 0) {
                    if (temp & 1) {

                        // Include numbers with the
                        // next digit as i
                        number = i * power(10, curLen)
                            + number;
                    }
                    else {

                        // Include numbers with the next
                        // next digit as j
                        number = j * power(10, curLen)
                            + number;
                    }

                    // Update temp
                    temp = (temp >> 1);
                    curLen++;
                }

                // Insert the current number into the set
                helper.add(number);
                while (curLen <= len) {
                    number = j * power(10, curLen) + number;
                    helper.add(number);
                    curLen++;
                }
            }
        }
    }

    // Loop to insert all the integers into
    // a vector from the set if the unique digits
    // in the integer is exactly two.
    for (let cur of helper) {

        // Stores the unique digits
        const count = new Set();
        let orz = cur;
        while (cur > 0) {
            count.add(cur % 10);
            cur = cur / 10;
        }

        // If count of exactly two
        if (count.size === 2) {
            nums.push(orz);
        }
    }
}

// Driver Code
    const arr = [123, 234];
    const N = arr.length;

    preProcess();
    nextGreaterEle(arr, N);

Output

131 242 

Time Complexity: O(10⁶ + N * log N)  = O(N * log N)
Auxiliary Space: O(10⁶) = O(1)