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Minimum and maximum number of digits required to be removed to make a given number divisible by 3
  • Last Updated : 04 Dec, 2020

Given a numeric string S, the task is to find the minimum and the maximum number of digits that must be removed from S such that it is divisible by 3. If it is impossible to do so, then print “-1”.

Examples:

Input: S = “12345”
Output:
Minimum: 0
Maximum: 4
Explanation: The given number 12345 is divisible by 3. Therefore, the minimum digits required to be removed to make it divisible by 3 is 0. After removing digits 1, 2, 4 and 5, only 3 remains, which is divisible by 3. Therefore, the maximum number of digits required to be removed is 4.

Input: S = “44”
Output:
Minimum: -1
Maximum: -1

Approach: The problem can be solved based on the observation that for any number to be divisible by 3, the sum of digits must also be divisible by 3. Follow the steps below to solve this problem:



  • Insert each digit of the given string into an array, say arr[], as (S[i] – ‘0’) % 3.
  • Find the count of 0s, 1s, and 2s in the array arr[].
  • Then, calculate the sum of digits, i.e. (arr[0] % 3) + (arr[1] % 3) + (arr[2] % 3)….. Update sum = sum % 3.
  • Depending on the value of sum, following three cases arise: 
    • If sum = 0: Minimum number of digits required to be removed is 0.
    • If sum = 1: Those digits should be removed whose sum gives a remainder 1 on dividing by 3. Therefore, following situations need to be considered: 
      • If the count of 1s is greater than or equal to 1, then the minimum number of digits required to be removed is 1.
      • If the count of 2s is greater than or equal to 2, then the minimum number of digits required to be removed will be 2 [Since (2 + 2) % 3 = 1].
    • If sum = 3: Those digits should be removed whose sum gives a remainder 2 on dividing by 3. Therefore, following situations arise: 
      • If the count of 2s is greater than or equal to 1, then the minimum number of digits required to be removed will be 1.
      • Otherwise, if the count of 1s is greater than or equal to 2, then the minimum number of digits to be removed will be 2 [(1 + 1) % 3 = 2].
  • For finding the maximum number of digits to be removed, following three cases need to be considered: 
    • If the count of 0s is greater than or equal to 1, then the maximum digits required to be removed will be (number of digits – 1).
    • If both the count of 1s and 2s are greater than or equal to 1, then maximum digits required to be removed will be (number of digits – 2) [(1+2) % 3 = 0].
    • If the count of either 1s or 2s is greater than or equal to 3, then maximum digits required to be removed will be (number of digits – 3) [(1+1+1) % 3 = 0, (2+2+2) % 3 = 0].

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum and
// minimum number of digits to be
// removed to make str divisible by 3
void minMaxDigits(string str, int N)
{
    // Convert the string into
    // array of digits
    int arr[N];
    for (int i = 0; i < N; i++)
        arr[i] = (str[i] - '0') % 3;
 
    // Count of 0s, 1s, and 2s
    int zero = 0, one = 0, two = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        if (arr[i] == 0)
            zero++;
        if (arr[i] == 1)
            one++;
        if (arr[i] == 2)
            two++;
    }
 
    // Find the sum of digits % 3
    int sum = 0;
 
    for (int i = 0; i < N; i++) {
        sum = (sum + arr[i]) % 3;
    }
 
    // Cases to find minimum number
    // of digits to be removed
    if (sum == 0) {
        cout << 0 << ' ';
    }
    if (sum == 1) {
        if (one && N > 1)
            cout << 1 << ' ';
        else if (two > 1 && N > 2)
            cout << 2 << ' ';
        else
            cout << -1 << ' ';
    }
    if (sum == 2) {
        if (two && N > 1)
            cout << 1 << ' ';
        else if (one > 1 && N > 2)
            cout << 2 << ' ';
        else
            cout << -1 << ' ';
    }
 
    // Cases to find maximum number
    // of digits to be removed
    if (zero > 0)
        cout << N - 1 << ' ';
    else if (one > 0 && two > 0)
        cout << N - 2 << ' ';
    else if (one > 2 || two > 2)
        cout << N - 3 << ' ';
    else
        cout << -1 << ' ';
}
 
// Driver Code
int main()
{
    string str = "12345";
    int N = str.length();
 
    // Function Call
    minMaxDigits(str, N);
 
    return 0;
}

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Java

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// Java program for the above approach
class GFG{
      
// Function to find the maximum and
// minimum number of digits to be
// removed to make str divisible by 3
static void minMaxDigits(String str, int N)
{
     
    // Convert the string into
    // array of digits
    int arr[] = new int[N];
    for(int i = 0; i < N; i++)
        arr[i] = (str.charAt(i) - '0') % 3;
  
    // Count of 0s, 1s, and 2s
    int zero = 0, one = 0, two = 0;
  
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
        if (arr[i] == 0)
            zero++;
        if (arr[i] == 1)
            one++;
        if (arr[i] == 2)
            two++;
    }
  
    // Find the sum of digits % 3
    int sum = 0;
  
    for(int i = 0; i < N; i++)
    {
        sum = (sum + arr[i]) % 3;
    }
  
    // Cases to find minimum number
    // of digits to be removed
    if (sum == 0)
    {
        System.out.print(0 + " ");
    }
    if (sum == 1)
    {
        if ((one != 0) && (N > 1))
            System.out.print(1 + " ");
        else if (two > 1 && N > 2)
            System.out.print(2 + " ");
        else
            System.out.print(-1 + " ");
    }
    if (sum == 2)
    {
        if (two != 0 && N > 1)
            System.out.print(1 + " ");
        else if (one > 1 && N > 2)
            System.out.print(2 + " ");
        else
            System.out.print(-1 + " ");
    }
  
    // Cases to find maximum number
    // of digits to be removed
    if (zero > 0)
        System.out.print(N - 1 + " ");
    else if (one > 0 && two > 0)
        System.out.print(N - 2 + " ");
    else if (one > 2 || two > 2)
        System.out.print(N - 3 + " ");
    else
        System.out.print(-1 + " ");
}
  
// Driver code
public static void main(String[] args)
{
    String str = "12345";
    int N = str.length();
     
    // Function Call
    minMaxDigits(str, N);
}
}
  
// This code is contributed by sanjoy_62

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Python3

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# Python3 program for the above approach
 
# Function to find the maximum and
# minimum number of digits to be
# removed to make str divisible by 3
def minMaxDigits(str, N):
     
    # Convert the string into
    # array of digits
    arr = [0]* N
     
    for i in range(N):
        arr[i] = (ord(str[i]) -
                  ord('0')) % 3
  
    # Count of 0s, 1s, and 2s
    zero = 0
    one = 0
    two = 0
  
    # Traverse the array
    for i in range(N):
        if (arr[i] == 0):
            zero += 1
        if (arr[i] == 1):
            one += 1
        if (arr[i] == 2):
            two += 1
             
    # Find the sum of digits % 3
    sum = 0
  
    for i in range(N):
        sum = (sum + arr[i]) % 3
     
    # Cases to find minimum number
    # of digits to be removed
    if (sum == 0):
        print("0", end = " ")
     
    if (sum == 1):
        if (one and N > 1):
            print("1", end = " ")
        elif (two > 1 and N > 2):
            print("2", end = " ")
        else:
            print("-1", end = " ")
     
    if (sum == 2):
        if (two and N > 1):
            print("1", end = " ")
        elif (one > 1 and N > 2):
            print("2", end = " ")
        else:
            print("-1", end = " ")
     
    # Cases to find maximum number
    # of digits to be removed
    if (zero > 0):
        print(N - 1, end = " ")
    elif (one > 0 and two > 0):
        print(N - 2, end = " ")
    elif (one > 2 or two > 2):
        print(N - 3, end = " ")
    else :
        print("-1", end = " ")
 
# Driver Code
str = "12345"
N = len(str)
  
# Function Call
minMaxDigits(str, N)
 
# This code is contributed by susmitakundugoaldanga

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C#

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// C# program for the above approach
using System;
 
class GFG{
       
// Function to find the maximum and
// minimum number of digits to be
// removed to make str divisible by 3
static void minMaxDigits(string str, int N)
{
     
    // Convert the string into
    // array of digits
    int[] arr = new int[N];
    for(int i = 0; i < N; i++)
        arr[i] = (str[i] - '0') % 3;
         
    // Count of 0s, 1s, and 2s
    int zero = 0, one = 0, two = 0;
   
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
        if (arr[i] == 0)
            zero++;
        if (arr[i] == 1)
            one++;
        if (arr[i] == 2)
            two++;
    }
   
    // Find the sum of digits % 3
    int sum = 0;
   
    for(int i = 0; i < N; i++)
    {
        sum = (sum + arr[i]) % 3;
    }
   
    // Cases to find minimum number
    // of digits to be removed
    if (sum == 0)
    {
        Console.Write(0 + " ");
    }
    if (sum == 1)
    {
        if ((one != 0) && (N > 1))
            Console.Write(1 + " ");
        else if (two > 1 && N > 2)
            Console.Write(2 + " ");
        else
            Console.Write(-1 + " ");
    }
    if (sum == 2)
    {
        if (two != 0 && N > 1)
            Console.Write(1 + " ");
        else if (one > 1 && N > 2)
            Console.Write(2 + " ");
        else
            Console.Write(-1 + " ");
    }
   
    // Cases to find maximum number
    // of digits to be removed
    if (zero > 0)
        Console.Write(N - 1 + " ");
    else if (one > 0 && two > 0)
        Console.Write(N - 2 + " ");
    else if (one > 2 || two > 2)
        Console.Write(N - 3 + " ");
    else
        Console.Write(-1 + " ");
}
   
// Driver code
public static void Main()
{
    string str = "12345";
    int N = str.Length;
      
    // Function Call
    minMaxDigits(str, N);
}
}
 
// This code is contributed by code_hunt

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Output: 

0 4

 

Time Complexity: O(log10N)
Auxiliary Space: O(log10N)

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