Given a string of consecutive digits and a number Y, the task is to find the number of minimum sets such that every set follows the below rule:
- Set should contain consecutive numbers
- No digit can be used more than once.
- The number in the set should not be more than Y.
Examples:
Input: s = "1234", Y = 30
Output: 3
Three sets of {12, 3, 4}
Input: s = "1234", Y = 4
Output: 4
Four sets of {1}, {2}, {3}, {4}
Approach: The following problem can be solved using a greedy approach whose steps are given below:
- Iterate in the string, and concatenate the number to a variable(let say num) using num*10 + (s[i]-'0')
- If the number is not greater than Y, then mark f = 1.
- If the number exceeds Y, then increase count if f = 1 and re-initialize f as 0 and also initialize num as s[i]-'0' or num as 0.
- After iterating in the string completely, then increase the count if f is 1.
Below is the implementation of the above approach:
// C++ program to find the minimum number
// sets with consecutive numbers and less than Y
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum number of shets
int minimumSets(string s, int y)
{
// Variable to count
// the number of sets
int cnt = 0;
int num = 0;
int l = s.length();
int f = 0;
// Iterate in the string
for (int i = 0; i < l; i++) {
// Add the number to string
num = num * 10 + (s[i] - '0');
// Mark that we got a number
if (num <= y)
f = 1;
else // Every time it exceeds
{
// Check if previous was
// anytime less than Y
if (f)
cnt += 1;
// Current number
num = s[i] - '0';
f = 0;
// Check for current number
if (num <= y)
f = 1;
else
num = 0;
}
}
// Check for last added number
if (f)
cnt += 1;
return cnt;
}
// Driver Code
int main()
{
string s = "1234";
int y = 30;
cout << minimumSets(s, y);
return 0;
}
// Java program to find the minimum number
// sets with consecutive numbers and less than Y
import java.util.*;
class solution
{
// Function to find the minimum number of shets
static int minimumSets(String s, int y)
{
// Variable to count
// the number of sets
int cnt = 0;
int num = 0;
int l = s.length();
boolean f = false;
// Iterate in the string
for (int i = 0; i < l; i++) {
// Add the number to string
num = num * 10 + (s.charAt(i) - '0');
// Mark that we got a number
if (num <= y)
f = true;
else // Every time it exceeds
{
// Check if previous was
// anytime less than Y
if (f)
cnt += 1;
// Current number
num = s.charAt(i) - '0';
f = false;
// Check for current number
if (num <= y)
f = true;
else
num = 0;
}
}
// Check for last added number
if (f == true)
cnt += 1;
return cnt;
}
// Driver Code
public static void main(String args[])
{
String s = "1234";
int y = 30;
System.out.println(minimumSets(s, y));
}
}
// This code is contributed by
// Shashank_Sharma
# Python3 program to find the minimum number
# sets with consecutive numbers and less than Y
import math as mt
# Function to find the minimum number of shets
def minimumSets(s, y):
# Variable to count the number of sets
cnt = 0
num = 0
l = len(s)
f = 0
# Iterate in the string
for i in range(l):
# Add the number to string
num = num * 10 + (ord(s[i]) - ord('0'))
# Mark that we got a number
if (num <= y):
f = 1
else: # Every time it exceeds
# Check if previous was anytime
# less than Y
if (f):
cnt += 1
# Current number
num = ord(s[i]) - ord('0')
f = 0
# Check for current number
if (num <= y):
f = 1
else:
num = 0
# Check for last added number
if (f):
cnt += 1
return cnt
# Driver Code
s = "1234"
y = 30
print(minimumSets(s, y))
# This code is contributed by
# Mohit kumar 29
// C# program to find the minimum number
// sets with consecutive numbers and less than Y
using System;
class solution
{
// Function to find the minimum number of shets
static int minimumSets(string s, int y)
{
// Variable to count
// the number of sets
int cnt = 0;
int num = 0;
int l = s.Length ;
bool f = false;
// Iterate in the string
for (int i = 0; i < l; i++) {
// Add the number to string
num = num * 10 + (s[i] - '0');
// Mark that we got a number
if (num <= y)
f = true;
else // Every time it exceeds
{
// Check if previous was
// anytime less than Y
if (f)
cnt += 1;
// Current number
num = s[i] - '0';
f = false;
// Check for current number
if (num <= y)
f = true;
else
num = 0;
}
}
// Check for last added number
if (f == true)
cnt += 1;
return cnt;
}
// Driver Code
public static void Main()
{
string s = "1234";
int y = 30;
Console.WriteLine(minimumSets(s, y));
}
// This code is contributed by Ryuga
}
<?php
// PHP program to find the minimum number
// sets with consecutive numbers and less than Y
// Function to find the minimum number of shets
function minimumSets($s, $y)
{
// Variable to count
// the number of sets
$cnt = 0;
$num = 0;
$l = strlen($s);
$f = 0;
// Iterate in the string
for ($i = 0; $i < $l; $i++)
{
// Add the number to string
$num = $num * 10 + ($s[$i] - '0');
// Mark that we got a number
if ($num <= $y)
$f = 1;
else // Every time it exceeds
{
// Check if previous was
// anytime less than Y
if ($f)
$cnt += 1;
// Current number
$num = $s[$i] - '0';
$f = 0;
// Check for current number
if ($num <= $y)
$f = 1;
else
$num = 0;
}
}
// Check for last added number
if ($f)
$cnt += 1;
return $cnt;
}
// Driver Code
$s = "1234";
$y = 30;
echo(minimumSets($s, $y));
//This code is contributed by Shivi_Aggarwal
?>
<script>
// Javascript program to find the minimum number
// sets with consecutive numbers and less than Y
// Function to find the minimum number of shets
function minimumSets(s, y)
{
// Variable to count
// the number of sets
let cnt = 0;
let num = 0;
let l = s.length;
let f = false;
// Iterate in the string
for (let i = 0; i < l; i++) {
// Add the number to string
num = num * 10 + (s[i] - '0');
// Mark that we got a number
if (num <= y)
f = true;
else // Every time it exceeds
{
// Check if previous was
// anytime less than Y
if (f)
cnt += 1;
// Current number
num = s[i] - '0';
f = false;
// Check for current number
if (num <= y)
f = true;
else
num = 0;
}
}
// Check for last added number
if (f == true)
cnt += 1;
return cnt;
}
// Driver code
let s = "1234";
let y = 30;
document.write(minimumSets(s, y));
</script>
Output:
3
Time Complexity: O(N), as we are using a loop to traverse N times. Where N is the length of the string.
Auxiliary Space: O(1), as we are not using any extra space.
