Reduce a given number to form a key by the given operations
Last Updated :
22 Apr, 2021
Given an integer N, the task is to reduce the number and form a key by following the operations given below:
- Extract the Most Significant Digit of the number:
- If the digit is even: Add the consecutive digits until the sum of digits is odd.
- If the digit is odd: Add the consecutive digits until the sum of digits is even.
- Repeat the process for all the remaining digits.
- Finally, concatenate the sum computed together to get the key.
Examples:
Input: N = 1667848270
Output: 20290
Explanation:
Step 1: First Digit(= 1) is odd. So, add up the next digits until the sum is even.
Therefore, digits 1, 6, 6, and 7 are added up to form 20.
Step 2: Next digit(= 8) is even. So, add up the next digits until the sum is odd.
Therefore, digits 8, 4, 8, 2, and 7 are added up to form 29.
Step 3: Last digit(= 0) is even.
Therefore, the final answer after concatenating the results will be: 20290
Input: N = 7246262412
Output: 342
Explanation:
Step 1: First Digit(= 7) is odd. So, add up the next digits until the sum is even.
Therefore, digits 7, 2, 4, 6, 2, 6, 2, 4, and 1 are added up to form 34.
Step 2: Last digit(= 2) is even.
Therefore, the final answer after concatenating the results will be: 342.
Approach: The idea is to iterate the digits of the number and check the parity of the digit. If it is even, then proceed to the next digits until an odd digit is encountered. For odd digit, add consecutive digits until the sum of digits is even. Finally, concatenate the sum computed to get the desired key.
Below is the implementation of the above approach:
C++
// C++ program of the
// above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the key
// of the given number
int key(int N)
{
// Convert the integer
// to String
string num = "" + to_string(N);
int ans = 0;
int j = 0;
// Iterate the num-string
// to get the result
for(j = 0; j < num.length(); j++)
{
// Check if digit is even or odd
if ((num[j] - 48) % 2 == 0)
{
int add = 0;
int i;
// Iterate until odd sum
// is obtained by adding
// consecutive digits
for(i = j; j < num.length(); j++)
{
add += num[j] - 48;
// Check if sum becomes odd
if (add % 2 == 1)
break;
}
if (add == 0)
{
ans *= 10;
}
else
{
int digit = (int)floor(log10(add) + 1);
ans *= (pow(10, digit));
// Add the result in ans
ans += add;
}
// Assign the digit index
// to num string
i = j;
}
else
{
// If the number is odd
int add = 0;
int i;
// Iterate until odd sum
// is obtained by adding
// consecutive digits
for(i = j; j < num.length(); j++)
{
add += num[j] - 48;
// Check if sum becomes even
if (add % 2 == 0)
{
break;
}
}
if (add == 0)
{
ans *= 10;
}
else
{
int digit = (int)floor(
log10(add) + 1);
ans *= (pow(10, digit));
// Add the result in ans
ans += add;
}
// assign the digit index
// to main numstring
i = j;
}
}
// Check if all digits
// are visited or not
if (j + 1 >= num.length())
{
return ans;
}
else
{
return ans += num[num.length() - 1] - 48;
}
}
// Driver code
int main()
{
int N = 1667848271;
cout << key(N);
return 0;
}
// This code is contributed by divyeshrabadiya07
// Java program of the
// above approach
import java.io.*;
import java.util.*;
import java.lang.*;
public class Main {
// Function to find the key
// of the given number
static int key(int N)
{
// Convert the integer
// to String
String num = "" + N;
int ans = 0;
int j = 0;
// Iterate the num-string
// to get the result
for (j = 0; j < num.length(); j++) {
// Check if digit is even or odd
if ((num.charAt(j) - 48) % 2 == 0) {
int add = 0;
int i;
// Iterate until odd sum
// is obtained by adding
// consecutive digits
for (i = j; j < num.length(); j++) {
add += num.charAt(j) - 48;
// Check if sum becomes odd
if (add % 2 == 1)
break;
}
if (add == 0) {
ans *= 10;
}
else {
int digit = (int)Math.floor(
Math.log10(add) + 1);
ans *= (Math.pow(10, digit));
// Add the result in ans
ans += add;
}
// Assign the digit index
// to num string
i = j;
}
else {
// If the number is odd
int add = 0;
int i;
// Iterate until odd sum
// is obtained by adding
// consecutive digits
for (i = j; j < num.length(); j++) {
add += num.charAt(j) - 48;
// Check if sum becomes even
if (add % 2 == 0) {
break;
}
}
if (add == 0) {
ans *= 10;
}
else {
int digit = (int)Math.floor(
Math.log10(add) + 1);
ans *= (Math.pow(10, digit));
// Add the result in ans
ans += add;
}
// assign the digit index
// to main numstring
i = j;
}
}
// Check if all digits
// are visited or not
if (j + 1 >= num.length()) {
return ans;
}
else {
return ans += num.charAt(
num.length() - 1)
- 48;
}
}
// Driver Code
public static void main(String[] args)
{
int N = 1667848271;
System.out.print(key(N));
}
}
# Python3 program of the
# above approach
import math
# Function to find the key
# of the given number
def key(N) :
# Convert the integer
# to String
num = "" + str(N)
ans = 0
j = 0
# Iterate the num-string
# to get the result
while j < len(num) :
# Check if digit is even or odd
if ((ord(num[j]) - 48) % 2 == 0) :
add = 0
# Iterate until odd sum
# is obtained by adding
# consecutive digits
i = j
while j < len(num) :
add += ord(num[j]) - 48
# Check if sum becomes odd
if (add % 2 == 1) :
break
j += 1
if (add == 0) :
ans *= 10
else :
digit = int(math.floor(math.log10(add) + 1))
ans *= (pow(10, digit))
# Add the result in ans
ans += add
# Assign the digit index
# to num string
i = j
else :
# If the number is odd
add = 0
# Iterate until odd sum
# is obtained by adding
# consecutive digits
i = j
while j < len(num) :
add += ord(num[j]) - 48
# Check if sum becomes even
if (add % 2 == 0) :
break
j += 1
if (add == 0) :
ans *= 10
else :
digit = int(math.floor(math.log10(add) + 1))
ans *= (pow(10, digit))
# Add the result in ans
ans += add
# assign the digit index
# to main numstring
i = j
j += 1
# Check if all digits
# are visited or not
if (j + 1) >= len(num) :
return ans
else :
ans += ord(num[len(num) - 1]) - 48
return ans
N = 1667848271
print(key(N))
# This code is contributed by divyesh072019
// C# program of the
// above approach
using System;
class GFG{
// Function to find the key
// of the given number
static int key(int N)
{
// Convert the integer
// to String
String num = "" + N;
int ans = 0;
int j = 0;
// Iterate the num-string
// to get the result
for (j = 0; j < num.Length; j++)
{
// Check if digit is even or odd
if ((num[j] - 48) % 2 == 0)
{
int add = 0;
int i;
// Iterate until odd sum
// is obtained by adding
// consecutive digits
for (i = j; j < num.Length; j++)
{
add += num[j] - 48;
// Check if sum becomes odd
if (add % 2 == 1)
break;
}
if (add == 0)
{
ans *= 10;
}
else
{
int digit = (int)Math.Floor(
Math.Log10(add) + 1);
ans *= (int)(Math.Pow(10, digit));
// Add the result in ans
ans += add;
}
// Assign the digit index
// to num string
i = j;
}
else
{
// If the number is odd
int add = 0;
int i;
// Iterate until odd sum
// is obtained by adding
// consecutive digits
for (i = j; j < num.Length; j++)
{
add += num[j] - 48;
// Check if sum becomes even
if (add % 2 == 0)
{
break;
}
}
if (add == 0)
{
ans *= 10;
}
else {
int digit = (int)Math.Floor(
Math.Log10(add) + 1);
ans *= (int)(Math.Pow(10, digit));
// Add the result in ans
ans += add;
}
// assign the digit index
// to main numstring
i = j;
}
}
// Check if all digits
// are visited or not
if (j + 1 >= num.Length)
{
return ans;
}
else
{
return ans += num[num.Length - 1] - 48;
}
}
// Driver Code
public static void Main(String[] args)
{
int N = 1667848271;
Console.Write(key(N));
}
}
// This code is contributed by Rajput-Ji
<script>
// Javascript program of the above approach
// Function to find the key
// of the given number
function key(N)
{
// Convert the integer
// to String
let num = "" + N.toString();
let ans = 0;
let j = 0;
// Iterate the num-string
// to get the result
for (j = 0; j < num.length; j++)
{
// Check if digit is even or odd
if ((num[j].charCodeAt() - 48) % 2 == 0)
{
let add = 0;
let i;
// Iterate until odd sum
// is obtained by adding
// consecutive digits
for (i = j; j < num.length; j++)
{
add += num[j].charCodeAt() - 48;
// Check if sum becomes odd
if (add % 2 == 1)
break;
}
if (add == 0)
{
ans *= 10;
}
else
{
let digit = Math.floor(Math.log10(add) + 1);
ans *= parseInt(Math.pow(10, digit), 10);
// Add the result in ans
ans += add;
}
// Assign the digit index
// to num string
i = j;
}
else
{
// If the number is odd
let add = 0;
let i;
// Iterate until odd sum
// is obtained by adding
// consecutive digits
for (i = j; j < num.length; j++)
{
add += num[j].charCodeAt() - 48;
// Check if sum becomes even
if (add % 2 == 0)
{
break;
}
}
if (add == 0)
{
ans *= 10;
}
else {
let digit = Math.floor(Math.log10(add) + 1);
ans *= parseInt(Math.pow(10, digit), 10);
// Add the result in ans
ans += add;
}
// assign the digit index
// to main numstring
i = j;
}
}
// Check if all digits
// are visited or not
if (j + 1 >= num.length)
{
return ans;
}
else
{
return ans += num[num.length - 1].charCodeAt() - 48;
}
}
let N = 1667848271;
document.write(key(N));
// This code is contributed by mukesh07.
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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