Given a Large number N ( number of digits in N can be up to 105). The task is to find the cuts required of a number such that maximum parts are divisible by 3.
Examples:
Input: N = 1269 Output: 3 Cut the number as 12|6|9. So, 12, 6, 9 are the three numbers which are divisible by 3. Input: N = 71 Output: 0 However, we make cuts there is no such number that is divisible by 3.
Approach:
Let’s calculate the values of the array res[0…n], where res[i] is the answer for the prefix of the length i. Obviously, res[0]:=0, since for the empty string (the prefix of the length 0) the answer is 0.
For i>0 one can find res[i] in the following way:
- Let’s look at the last digit of the prefix of length i. It has index i-1. Either it doesn’t belong to segment divisible by 3, or it belongs.
- If it doesn’t belong, it means last digit can’t be used, so res[i]=res[i-1]. If it belongs then find shortest s[j..i-1] that is divisible by 3 and try to update res[i] with the value res[j]+1.
- A number is divisible by 3, if and only if the sum of its digits is divisible by 3. So the task is to find the shortest suffix of s[0…i-1] with the sum of digits divisible by 3. If such suffix is s[j..i-1] then s[0..j-1] and s[0..i-1] have the same remainder of the sum of digits modulo 3.
- Let’s maintain remIndex[0..2]- an array of the length 3, where remIndex[r] is the length of the longest processed prefix with the sum of digits equal to r modulo 3. Use remIndex[r]= -1 if there is no such prefix. It is easy to see that j=remIndex[r] where r is the sum of digits on the ith prefix modulo 3.
- So to find the maximal j<=i-1 that substring s[j..i-1] is divisible by 3, just check that remIndex[r] not equals to -1 and use j=remIndex[r], where r is the sum of digits on the i-th prefix modulo 3.
- It means that to handle case that the last digit belongs to divisible by 3 segment, try to update res[i] with value res[remIndex[r]]+1. In other words, just do if (remIndex[r] != -1) => res[i] = max(res[i], res[remIndex[r]] + 1).
Below is the implementation of the above approach:
C++
// CPP program to find the maximum number of// numbers divisible by 3 in a large number#include <bits/stdc++.h>using namespace std;// Function to find the maximum number of// numbers divisible by 3 in a large numberint MaximumNumbers(string s){ // store size of the string int n = s.length(); // Stores last index of a remainder vector<int> remIndex(3, -1); // last visited place of remainder // zero is at 0. remIndex[0] = 0; // To store result from 0 to i vector<int> res(n + 1); int r = 0; for (int i = 1; i <= n; i++) { // get the remainder r = (r + s[i-1] - '0') % 3; // Get maximum res[i] value res[i] = res[i-1]; if (remIndex[r] != -1) res[i] = max(res[i], res[remIndex[r]] + 1); remIndex[r] = i+1; } return res[n];}// Driver Codeint main(){ string s = "12345"; cout << MaximumNumbers(s); return 0;} |
Java
// Java program to find the maximum number of// numbers divisible by 3 in a large numberimport java.util.*;class GFG{ // Function to find the maximum number of// numbers divisible by 3 in a large numberstatic int MaximumNumbers(String s){ // store size of the string int n = s.length(); // Stores last index of a remainder int [] remIndex={-1, -1, -1}; // last visited place of remainder // zero is at 0. remIndex[0] = 0; // To store result from 0 to i int[] res = new int[n + 1]; int r = 0; for (int i = 1; i <= n; i++) { // get the remainder r = (r + s.charAt(i-1) - '0') % 3; // Get maximum res[i] value res[i] = res[i - 1]; if (remIndex[r] != -1) res[i] = Math.max(res[i], res[remIndex[r]] + 1); remIndex[r] = i + 1; } return res[n];}// Driver Codepublic static void main (String[] args){ String s = "12345"; System.out.println(MaximumNumbers(s));}}// This code is contributed by // chandan_jnu |
Python3
# Python3 program to find the maximum # number of numbers divisible by 3 in # a large numberimport math as mt # Function to find the maximum number # of numbers divisible by 3 in a # large numberdef MaximumNumbers(string): # store size of the string n = len(string) # Stores last index of a remainder remIndex = [-1 for i in range(3)] # last visited place of remainder # zero is at 0. remIndex[0] = 0 # To store result from 0 to i res = [-1 for i in range(n + 1)] r = 0 for i in range(n + 1): # get the remainder r = (r + ord(string[i - 1]) - ord('0')) % 3 # Get maximum res[i] value res[i] = res[i - 1] if (remIndex[r] != -1): res[i] = max(res[i], res[remIndex[r]] + 1) remIndex[r] = i + 1 return res[n]# Driver Codes= "12345"print(MaximumNumbers(s))# This code is contributed# by Mohit kumar 29 |
C#
// C# program to find the maximum number of // numbers divisible by 3 in a large number . using System;class GFG { // Function to find the maximum number of // numbers divisible by 3 in a large number static int MaximumNumbers(String s) { // store size of the string int n = s.Length; // Stores last index of a remainder int [] remIndex = {-1, -1, -1}; // last visited place of remainder // zero is at 0. remIndex[0] = 0; // To store result from 0 to i int[] res = new int[n + 1]; int r = 0; for (int i = 1; i <= n; i++) { // get the remainder r = (r + s[i-1] - '0') % 3; // Get maximum res[i] value res[i] = res[i - 1]; if (remIndex[r] != -1) res[i] = Math.Max(res[i], res[remIndex[r]] + 1); remIndex[r] = i + 1; } return res[n]; } // Driver Code public static void Main (String[] args) { String s = "12345"; Console.WriteLine(MaximumNumbers(s)); } } // This code has been contributed by // PrinciRaj1992 |
PHP
<?php// PHP program to find the maximum number of // numbers divisible by 3 in a large number // Function to find the maximum number of // numbers divisible by 3 in a large number function MaximumNumbers($s) { // store size of the string $n = strlen($s) ; // Stores last index of a remainder $remIndex = array_fill(0,3,-1) ; // last visited place of remainder // zero is at 0. $remIndex[0] = 0; // To store result from 0 to i $res = array() ; $r = 0; for ($i = 1; $i <= $n; $i++) { // get the remainder $r = ($r + $s[$i-1] - '0') % 3; // Get maximum res[i] value $res[$i] = $res[$i-1]; if ($remIndex[$r] != -1) $res[$i] = max($res[$i], $res[$remIndex[$r]] + 1); $remIndex[$r] = $i+1; } return $res[$n]; } // Driver Code $s = "12345"; print(MaximumNumbers($s)) # This code is contributed by Ryuga?> |
Javascript
<script>// Javascript program to find the maximum number of// numbers divisible by 3 in a large number// Function to find the maximum number of// numbers divisible by 3 in a large numberfunction MaximumNumbers(s){ // store size of the string let n = s.length; // Stores last index of a remainder let remIndex=[-1, -1, -1]; // last visited place of remainder // zero is at 0. remIndex[0] = 0; // To store result from 0 to i let res = new Array(n + 1); for(let i=0;i<res.length;i++) { res[i]=0; } let r = 0; for (let i = 1; i <= n; i++) { // get the remainder r = (r + s[i-1].charCodeAt(0) - '0'.charCodeAt(0)) % 3; // Get maximum res[i] value res[i] = res[i - 1]; if (remIndex[r] != -1) res[i] = Math.max(res[i], res[remIndex[r]] + 1); remIndex[r] = i + 1; } return res[n];}// Driver Codelet s = "12345";document.write(MaximumNumbers(s));// This code is contributed by patel2127</script> |
Output:
3
Time Complexity: O(n), since there runs a loop from 1 to n.
Auxiliary Space: O(n), since extra space has been taken in the form of an array of size of n the space takes in linear.
Another Approach:
We can use running_sum which keeps the sum of all successive integers, where none of the individual integers is divisible by 3. we can pass through each integer one by one and do the following:
- If the integer is divisible by 3 or the running_sum is non-zero and divisible by 3, increment the counter and reset running_sum.
- In case the integer is not divisible by 3, keep a track of sum of all such successive integers.
C++
// C++ program to find the maximum number// of numbers divisible by 3 in large number #include <iostream>using namespace std;int get_max_splits(string num_string){ // This will contain the count of the splits int count = 0, current_num; // This will keep sum of all successive // integers, when they are indivisible by 3 int running_sum = 0; for (int i = 0; i < num_string.length(); i++) { current_num = num_string[i] - '0'; running_sum += current_num; // This is the condition of finding a split if (current_num % 3 == 0 || (running_sum != 0 && running_sum % 3 == 0)) { count += 1; running_sum = 0; } } return count;}// Driver codeint main(){ cout << get_max_splits("12345") << endl; return 0;}// This code is contributed by Rituraj Jain |
Java
// Java program to find the maximum number// of numbers divisible by 3 in large number class GFG{static int get_max_splits(String num_String){ // This will contain the count of the splits int count = 0, current_num; // This will keep sum of all successive // integers, when they are indivisible by 3 int running_sum = 0; for (int i = 0; i < num_String.length(); i++) { current_num = num_String.charAt(i) - '0'; running_sum += current_num; // This is the condition of finding a split if (current_num % 3 == 0 || (running_sum != 0 && running_sum % 3 == 0)) { count += 1; running_sum = 0; } } return count;}// Driver codepublic static void main(String[] args){ System.out.print(get_max_splits("12345") +"\n"); }}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to find the maximum # number of numbers divisible by 3 in # a large numberdef get_max_splits(num_string): # This will contain the count of the splits count = 0 # This will keep sum of all successive # integers, when they are indivisible by 3 running_sum = 0 for i in range(len(num_string)): current_num = int(num_string[i]) running_sum += current_num # This is the condition of finding a split if current_num % 3 == 0 or (running_sum != 0 and running_sum % 3 == 0): count += 1 running_sum = 0 return countprint get_max_splits("12345")# This code is contributed by Amit Ranjan |
C#
// C# program to find the maximum number// of numbers divisible by 3 in large number using System;class GFG{static int get_max_splits(String num_String){ // This will contain the count of the splits int count = 0, current_num; // This will keep sum of all successive // integers, when they are indivisible by 3 int running_sum = 0; for (int i = 0; i < num_String.Length; i++) { current_num = num_String[i] - '0'; running_sum += current_num; // This is the condition of finding a split if (current_num % 3 == 0 || (running_sum != 0 && running_sum % 3 == 0)) { count += 1; running_sum = 0; } } return count;}// Driver codepublic static void Main(String[] args){ Console.Write(get_max_splits("12345") +"\n"); }}// This code is contributed by 29AjayKumar |
PHP
<?php// PHP program to find the maximum // number of numbers divisible by 3 in // a large numberfunction get_max_splits($num_string){ // This will contain the count of // the splits $count = 0; // This will keep sum of all successive // integers, when they are indivisible by 3 $running_sum = 0; for ($i = 0; $i < strlen($num_string); $i++) { $current_num = intval($num_string[$i]); $running_sum += $current_num; // This is the condition of finding a split if ($current_num % 3 == 0 or ($running_sum != 0 and $running_sum % 3 == 0)) { $count += 1; $running_sum = 0; } } return $count;}// Driver Codeprint(get_max_splits("12345"));// This code is contributed by mits?> |
Javascript
<script>// JavaScript program to find the maximum number// of numbers divisible by 3 in large numberfunction get_max_splits(num_String){ // This will contain the count of the splits let count = 0, current_num; // This will keep sum of all successive // integers, when they are indivisible by 3 let running_sum = 0; for (let i = 0; i < num_String.length; i++) { current_num = num_String[i].charCodeAt(0) - '0'.charCodeAt(0); running_sum += current_num; // This is the condition of finding a split if (current_num % 3 == 0 || (running_sum != 0 && running_sum % 3 == 0)) { count += 1; running_sum = 0; } } return count;}// Driver codedocument.write(get_max_splits("12345") +"<br>");// This code is contributed by unknown2108</script> |
Output:
3
Time Complexity: O(n) only one traversal of the array is needed so the algorithm takes overall linear time
Auxiliary Space: O(1) since no extra array is used so it takes constant space
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
