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Least Greater number with same digit sum
  • Difficulty Level : Hard
  • Last Updated : 02 May, 2019

Given a number represented in the form of an array such that each element of the array stores a single digit of the number. That is, array for the number 1234 will be arr[] = {1,2,3,4}. The task is to find the least number greater than the given number but having the sum of digits equal to the given number.

For simplicity: Consider the length of number can be 20 at maximum.

Examples:

Input : arr[] = {0, 0, 0, 0, 0, 0, 0, 3, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 8 };
Output : 00000004799999999999
Explanation : Sum of digits = 110

Input : arr[] = {0, 0, 0, 0, 0, 0, 0, 3, 9, 7, 0, 0, 2, 9, 8, 9, 5, 9, 9, 0};
Output : 00000003970029896089
Explanation : Sum of digits = 70



A Brute Force Approach is to:

  1. Start from that number and increment the number by one.
  2. Check the sum. If the sum is same the return the number.
  3. Else return to step one.

A Better Approach is to jump to the next number in O(n) time complexity, where n is the length of string.

We divide the number into 4 regions :
1st: Trailing zeros .
2nd: The lowest digit not in Region 1.
3rd: Consecutive 9s starting with the digit above Region 2.
4th: All remaining digits.

Then the next number is :
[Region 4+1] [Region 1] [Region 2-1] [Region 3] .

Example:
Input Number = 548995000
Region 1 : 000
Region 2 : 5
Region 3 : 99
Region 4 : 548

Next number = 549000499

Below is the implementation of the above approach:

C++

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// CPP program to find next greater number with
// same sum of digits.
#include <bits/stdc++.h>
using namespace std;
  
#define pb push_back
  
void getnext(int arr[], int n)
{
    // for storing 4 regions
    vector<int> a1, a2, a3, a4;
  
    // trailing zeros region1
    int i = n - 1; // last index
    while (arr[i] == 0) 
    {
        a1.pb(0);
        i--;
    }
  
    // lowest region(region 2) not in region 1
    a2.pb(arr[i--]); 
  
    // Consecutive 9's   (region 3)
    while (arr[i] == 9) 
    {
        a3.pb(9);
        i--;
    }
  
    int j = 0;
    while (arr[j] == 0)
        j++; // Starting zeros
  
    while (j <= i) // 4th region
    {
        a4.pb(arr[j]);
        j++;
    }
  
    // Calculation of result
    j = a4.size() - 1;
  
    a4[j]++; // Region4 + 1
  
    a2[0]--; // Region2 -1
  
    int l = a1.size() + a2.size() + a3.size() + a4.size();
  
    // Calculating the result
    j = n-1;
    i = a3.size() - 1;
  
    // Copying the third region
    while (i >= 0)
    {
        arr[j--] = a3[i--];
    }
  
    // Copying the 2nd region
    i = a2.size() - 1;
    while (i >= 0) 
    {
        arr[j--] = a2[i--];
    }
  
    // Copying the 1st region
    i = a1.size() - 1;
    while (i >= 0) 
    {
        arr[j--] = a1[i--];
    }
  
    // Copying the 4th region
    i = a4.size() - 1;
    while (i >= 0) 
    {
        arr[j--] = a4[i--];
    }
  
    while (j >= 0)
        arr[j--] = 0;
}
  
int main()
{
    int arr[] = { 0, 0, 0, 0, 0, 0, 0, 3, 9, 7, 
                0, 0, 2, 9, 8, 9, 5, 9, 9, 0 };
    int n = sizeof(arr)/sizeof(arr[0]);
  
    getnext(arr, n); // Calling the function
  
    for (int i = 0; i < n; i++)
        cout << arr[i];
  
    return 0;
}

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Java

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// Java program to find next greater number with
// same sum of digits.
import java.util.*;
  
class GFG
{
  
static void getnext(int []arr, int n)
{
    // for storing 4 regions
    ArrayList<Integer> a1 = new ArrayList<Integer>();
    ArrayList<Integer> a2 = new ArrayList<Integer>();
    ArrayList<Integer> a3 = new ArrayList<Integer>();
    ArrayList<Integer> a4 = new ArrayList<Integer>();
  
    // trailing zeros region1
    int i = n - 1; // last index
    while (arr[i] == 0
    {
        a1.add(0);
        i--;
    }
  
    // lowest region(region 2) not in region 1
    a2.add(arr[i--]); 
  
    // Consecutive 9's (region 3)
    while (arr[i] == 9
    {
        a3.add(9);
        i--;
    }
  
    int j = 0;
    while (arr[j] == 0)
        j++; // Starting zeros
  
    while (j <= i) // 4th region
    {
        a4.add(arr[j]);
        j++;
    }
  
    // Calculation of result
    j = a4.size() - 1;
  
    a4.set(j,a4.get(j) + 1); // Region4 + 1
  
    a2.set(0,a2.get(0) - 1); // Region2 -1
  
    //int l = a1.size() + a2.size() + a3.size() + a4.size();
  
    // Calculating the result
    j = n - 1;
    i = a3.size() - 1;
  
    // Copying the third region
    while (i >= 0)
    {
        arr[j--] = (int)a3.get(i);
        i--;
    }
  
    // Copying the 2nd region
    i = a2.size() - 1;
    while (i >= 0
    {
        arr[j--] = (int)a2.get(i);
        i--;
    }
  
    // Copying the 1st region
    i = a1.size() - 1;
    while (i >= 0
    {
        arr[j--] = a1.get(i);
        i--;
    }
  
    // Copying the 4th region
    i = a4.size() - 1;
    while (i >= 0
    {
        arr[j--] = a4.get(i);
        i--;
    }
  
    while (j >= 0)
        arr[j--] = 0;
}
  
// Driver code
public static void main (String[] args) 
{
    int []arr = { 0, 0, 0, 0, 0, 0, 0, 3, 9, 7
                0, 0, 2, 9, 8, 9, 5, 9, 9, 0 };
    int n = arr.length;
  
    getnext(arr, n); // Calling the function
  
    for (int i = 0; i < n; i++)
        System.out.print(arr[i]);
}
}
  
// This code is contributed by mits

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Python3

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# Python3 program to find next greater number with
# same sum of digits.
  
def getnext(arr, n):
  
    # for storing 4 regions
    a1=[];
    a2=[];
    a3=[];
    a4=[];
  
    # trailing zeros region1
    i = n - 1; # last index
    while (arr[i] == 0):
        a1.append(0);
        i-=1;
  
    # lowest region(region 2) not in region 1
    a2.append(arr[i]);
    i-=1;
  
    # Consecutive 9's (region 3)
    while (arr[i] == 9):
        a3.append(9);
        i-=1;
  
    j = 0;
    while (arr[j] == 0):
        j+=1; # Starting zeros
  
    while (j <= i): # 4th region
        a4.append(arr[j]);
        j+=1;
  
    # Calculation of result
    j = len(a4) - 1;
  
    a4[j]+=1; # Region4 + 1
  
    a2[0]-=1; # Region2 -1
  
    l = len(a1) + len(a2) + len(a3) + len(a4);
  
    # Calculating the result
    j = n-1;
    i = len(a3) - 1;
  
    # Copying the third region
    while (i >= 0):
        arr[j] = a3[i];
        j-=1;
        i-=1;
  
    # Copying the 2nd region
    i = len(a2) - 1;
    while (i >= 0):
        arr[j] = a2[i];
        j-=1;
        i-=1;
  
    # Copying the 1st region
    i = len(a1) - 1;
    while (i >= 0):
        arr[j] = a1[i];
        j-=1;
        i-=1;
  
    # Copying the 4th region
    i = len(a4) - 1;
    while (i >= 0):
        arr[j] = a4[i];
        j-=1;
        i-=1;
  
    while (j >= 0):
        arr[j] = 0;
        j-=1;
  
# Driver code
arr = [ 0, 0, 0, 0, 0, 0, 0, 3, 9, 7, 0,
            0, 2, 9, 8, 9, 5, 9, 9, 0 ];
n = len(arr);
  
getnext(arr, n); # Calling the function
  
for i in range(0,n):
    print(arr[i],end="");
  
# This code is contributed by mits

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C#

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// C# program to find next greater number with
// same sum of digits.
using System;
using System.Collections;
  
class GFG
{
  
static void getnext(int []arr, int n)
{
    // for storing 4 regions
    ArrayList a1 = new ArrayList();
    ArrayList a2 = new ArrayList();
    ArrayList a3 = new ArrayList();
    ArrayList a4 = new ArrayList();
  
    // trailing zeros region1
    int i = n - 1; // last index
    while (arr[i] == 0) 
    {
        a1.Add(0);
        i--;
    }
  
    // lowest region(region 2) not in region 1
    a2.Add(arr[i--]); 
  
    // Consecutive 9's (region 3)
    while (arr[i] == 9) 
    {
        a3.Add(9);
        i--;
    }
  
    int j = 0;
    while (arr[j] == 0)
        j++; // Starting zeros
  
    while (j <= i) // 4th region
    {
        a4.Add(arr[j]);
        j++;
    }
  
    // Calculation of result
    j = a4.Count - 1;
  
    a4[j] = (int)a4[j] + 1; // Region4 + 1
  
    a2[0] = (int)a2[0] - 1; // Region2 -1
  
    //int l = a1.Count + a2.Count + a3.Count + a4.Count;
  
    // Calculating the result
    j = n - 1;
    i = a3.Count - 1;
  
    // Copying the third region
    while (i >= 0)
    {
        arr[j--] = (int)a3[i];
        i--;
    }
  
    // Copying the 2nd region
    i = a2.Count - 1;
    while (i >= 0) 
    {
        arr[j--] = (int)a2[i];
        i--;
    }
  
    // Copying the 1st region
    i = a1.Count - 1;
    while (i >= 0) 
    {
        arr[j--] = (int)a1[i];
        i--;
    }
  
    // Copying the 4th region
    i = a4.Count - 1;
    while (i >= 0) 
    {
        arr[j--] = (int)a4[i];
        i--;
    }
  
    while (j >= 0)
        arr[j--] = 0;
}
  
// Driver code
static void Main()
{
    int []arr = { 0, 0, 0, 0, 0, 0, 0, 3, 9, 7, 
                0, 0, 2, 9, 8, 9, 5, 9, 9, 0 };
    int n = arr.Length;
  
    getnext(arr, n); // Calling the function
  
    for (int i = 0; i < n; i++)
        Console.Write(arr[i]);
}
}
  
// This code is contributed by mits

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PHP

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<?php
// PHP program to find next greater number with
// same sum of digits.
  
function getnext(&$arr, $n)
{
    // for storing 4 regions
    $a1=array();
    $a2=array();
    $a3=array();
    $a4=array();
  
    // trailing zeros region1
    $i = $n - 1; // last index
    while ($arr[$i] == 0) 
    {
        array_push($a1,0);
        $i--;
    }
  
    // lowest region(region 2) not in region 1
    array_push($a2,$arr[$i--]); 
  
    // Consecutive 9's (region 3)
    while ($arr[$i] == 9) 
    {
        array_push($a3,9);
        $i--;
    }
  
    $j = 0;
    while ($arr[$j] == 0)
        $j++; // Starting zeros
  
    while ($j <= $i) // 4th region
    {
        array_push($a4,$arr[$j]);
        $j++;
    }
  
    // Calculation of result
    $j = count($a4) - 1;
  
    $a4[$j]++; // Region4 + 1
  
    $a2[0]--; // Region2 -1
  
    $l = count($a1) + count($a2) + count($a3) + count($a4);
  
    // Calculating the result
    $j = $n-1;
    $i = count($a3) - 1;
  
    // Copying the third region
    while ($i >= 0)
    {
        $arr[$j--] = $a3[$i--];
    }
  
    // Copying the 2nd region
    $i = count($a2) - 1;
    while ($i >= 0) 
    {
        $arr[$j--] = $a2[$i--];
    }
  
    // Copying the 1st region
    $i = count($a1) - 1;
    while ($i >= 0) 
    {
        $arr[$j--] = $a1[$i--];
    }
  
    // Copying the 4th region
    $i = count($a4) - 1;
    while ($i >= 0) 
    {
        $arr[$j--] = $a4[$i--];
    }
  
    while ($j >= 0)
        $arr[$j--] = 0;
}
  
        // Driver code
    $arr = array( 0, 0, 0, 0, 0, 0, 0, 3, 9, 7, 
                0, 0, 2, 9, 8, 9, 5, 9, 9, 0 );
    $n = count($arr);
  
    getnext($arr, $n); // Calling the function
  
    for ($i = 0; $i < $n; $i++)
        echo $arr[$i];
  
// This code is contributed by mits
?>

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Output:

00000003970029896089

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