Append a digit in the end to make the number equal to the length of the remaining string
Last Updated :
04 Jun, 2022
Given a string str in which an integer is appended in the end (with or without leading zeroes). The task is to find a single digit from the range [0, 9] that must be appended in the end of the integer so that the number becomes equal to the length of remaining string. Print -1 if its not possible.
Examples:
Input: str = "geeksforgeeks1"
Output: 3
Length of "geeksforgeeks" is 13
So, 3 must be appended at the end of 1.
Input: str = "abcd0"
Output: 4
Approach: Find the number appended in the end of the string say num and append a 0 in the end which is the least digit possible i.e. num = num * 10. Now find the length of the remaining string ignoring the numeric from the end say len. Now the digit which must be appended will be digit = len - num. If digit is in the range [0, 9] then print it else print -1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the required digit
int find_digit(string s, int n)
{
// To store the position of the first
// numeric digit in the string
int first_digit = -1;
for (int i = n - 1; i >= 0; i--) {
if (s[i] < '0' || s[i] > '9') {
first_digit = i;
break;
}
}
first_digit++;
// To store the length of the
// string without the numeric
// digits in the end
int s_len = first_digit;
// pw stores the current power of 10
// and num is to store the number
// which is appended in the end
int num = 0, pw = 1;
int i = n - 1;
while (i >= 0) {
// If current character is
// a numeric digit
if (s[i] >= '0' && s[i] <= '9') {
// Get the current digit
int digit = s[i] - '0';
// Build the number
num = num + (pw * digit);
// If number exceeds the length
if (num >= s_len)
return -1;
// Next power of 10
pw = pw * 10;
}
i--;
}
// Append 0 in the end
num = num * 10;
// Required number that must be added
int req = s_len - num;
// If number is not a single digit
if (req > 9 || req < 0)
return -1;
return req;
}
// Driver code
int main()
{
string s = "abcd0";
int n = s.length();
cout << find_digit(s, n);
return 0;
}
// Java implementation of the approach
import java.io.*;
class GFG
{
// Function to return the required digit
static int find_digit(String s, int n)
{
// To store the position of the first
// numeric digit in the string
int first_digit = -1;
for (int i = n - 1; i >= 0; i--)
{
if (s.charAt(i) < '0' ||
s.charAt(i) > '9')
{
first_digit = i;
break;
}
}
first_digit++;
// To store the length of the
// string without the numeric
// digits in the end
int s_len = first_digit;
// pw stores the current power of 10
// and num is to store the number
// which is appended in the end
int num = 0, pw = 1;
int i = n - 1;
while (i >= 0)
{
// If current character is
// a numeric digit
if (s.charAt(i) >= '0' &&
s.charAt(i) <= '9')
{
// Get the current digit
int digit = s.charAt(i) - '0';
// Build the number
num = num + (pw * digit);
// If number exceeds the length
if (num >= s_len)
return -1;
// Next power of 10
pw = pw * 10;
}
i--;
}
// Append 0 in the end
num = num * 10;
// Required number that must be added
int req = s_len - num;
// If number is not a single digit
if (req > 9 || req < 0)
return -1;
return req;
}
// Driver code
public static void main (String[] args)
{
String s = "abcd0";
int n = s.length();
System.out.print(find_digit(s, n));
}
}
// This code is contributed by vt_m
# Python3 implementation of the approach
# Function to return the required digit
def find_digit(s, n):
# To store the position of the first
# numeric digit in the string
first_digit = -1
for i in range(n - 1, -1, -1):
if s[i] < '0' or s[i] > '9':
first_digit = i
break
first_digit += 1
# To store the length of the
# string without the numeric
# digits in the end
s_len = first_digit
num = 0
pw = 1
i = n - 1
while i >= 0:
# If current character is
# a numeric digit
if s[i] >= '0' and s[i] <= '9':
# Get the current digit
digit = ord(s[i]) - ord('0')
# Build the number
num = num + (pw * digit)
# If number exceeds the length
if num >= s_len:
return -1
# Next power of 10
pw = pw * 10
i -= 1
# Append 0 in the end
num = num * 10
# Required number that must be added
req = s_len - num
# If number is not a single digit
if req > 9 or req < 0:
return -1
return req
# Driver code
if __name__ == "__main__":
s = "abcd0"
n = len(s)
print(find_digit(s, n))
# This code is contributed by
# sanjeev2552
// C# implementation of the approach
using System;
class GFG
{
// Function to return the required digit
static int find_digit(String s, int n)
{
// To store the position of the first
// numeric digit in the string
int first_digit = -1, i;
for (i = n - 1; i >= 0; i--)
{
if (s[i] < '0' ||
s[i] > '9')
{
first_digit = i;
break;
}
}
first_digit++;
// To store the length of the
// string without the numeric
// digits in the end
int s_len = first_digit;
// pw stores the current power of 10
// and num is to store the number
// which is appended in the end
int num = 0, pw = 1;
i = n - 1;
while (i >= 0)
{
// If current character is
// a numeric digit
if (s[i] >= '0' &&
s[i] <= '9')
{
// Get the current digit
int digit = s[i] - '0';
// Build the number
num = num + (pw * digit);
// If number exceeds the length
if (num >= s_len)
return -1;
// Next power of 10
pw = pw * 10;
}
i--;
}
// Append 0 in the end
num = num * 10;
// Required number that must be added
int req = s_len - num;
// If number is not a single digit
if (req > 9 || req < 0)
return -1;
return req;
}
// Driver code
public static void Main (String[] args)
{
String s = "abcd0";
int n = s.Length;
Console.Write(find_digit(s, n));
}
}
// This code is contributed by PrinciRaj1992
<script>
// Javascript implementation of the approach
// Function to return the required digit
function find_digit(s, n)
{
// To store the position of the first
// numeric digit in the string
var first_digit = -1;
for (var i = n - 1; i >= 0; i--) {
if (s[i] < '0' || s[i] > '9') {
first_digit = i;
break;
}
}
first_digit++;
// To store the length of the
// string without the numeric
// digits in the end
var s_len = first_digit;
// pw stores the current power of 10
// and num is to store the number
// which is appended in the end
var num = 0, pw = 1;
var i = n - 1;
while (i >= 0) {
// If current character is
// a numeric digit
if (s[i] >= '0' && s[i] <= '9') {
// Get the current digit
var digit = s[i] - '0';
// Build the number
num = num + (pw * digit);
// If number exceeds the length
if (num >= s_len)
return -1;
// Next power of 10
pw = pw * 10;
}
i--;
}
// Append 0 in the end
num = num * 10;
// Required number that must be added
var req = s_len - num;
// If number is not a single digit
if (req > 9 || req < 0)
return -1;
return req;
}
// Driver code
var s = "abcd0";
var n = s.length;
document.write( find_digit(s, n));
</script>
Time Complexity: O(n)
Auxiliary Space: O(1)
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