Find smallest number with given digits and sum of digits
Last Updated :
24 Mar, 2023
Given two positive integers P and Q, find the minimum integer containing only digits P and Q such that the sum of the digits of the integer is N.
Example:
Input: N = 11, P = 4, Q = 7
Output: 47
Explanation: There are two possible integers that can be formed from 4 and 7 such that their sum is 11 i.e. 47 and 74. Since we need to find the minimum possible value, 47 is the required answer.
Input: N = 11, P = 9, Q = 7
Output: Not Possible
Explanation: It is not possible to create an integer using digits 7 and 9 such that their sum is 11.
Efficient Approach: Let's consider P is greater than or equal to Q, count_P denotes the number of occurrences of P and count_Q denoted the number of occurrences of Q in the resulting integer. So, the question can be represented in the form of an equation (P * count_P) + (Q * count_Q) = N, and in order to minimize the count of digits in the resulting integer count_P + count_Q should be as minimum as possible. It can be observed that since P >= Q, the maximum possible value of count_P that satisfies (P * count_P) + (Q * count_Q) = N will be the most optimal choice. Below are the steps for the above approach :
- Initialize count_P and count_Q as 0.
- If N is divisible by P, count_P = N/P and N=0.
- If N is not divisible by P, subtract Q from N and increment count_Q by 1.
- Repeat steps number 2 and 3 until N is greater than 0.
- If N != 0, it is not possible to generate the integer that satisfies the required conditions. Else the resulting integer will be count_Q times Q followed by count_P times P.
Below is the implementation of the above approach:
C++
// C++ Program of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the minimum
// integer having only digits P and
// Q and the sum of digits as N
void printMinInteger(int P, int Q, int N)
{
// If Q is greater that P then
// swap the values of P and Q
if (Q > P) {
swap(P, Q);
}
// If P and Q are both zero or
// if Q is zero and N is not
// divisible by P then there
// is no possible integer which
// satisfies the given conditions
if (Q == 0 && (P == 0 || N % P != 0)) {
cout << "Not Possible";
return;
}
int count_P = 0, count_Q = 0;
// Loop to find the maximum value
// of count_P that also satisfy
// P*count_P + Q*count_Q = N
while (N > 0) {
if (N % P == 0) {
count_P += N / P;
N = 0;
}
else {
N = N - Q;
count_Q++;
}
}
// If N is 0, there is a valid
// integer possible that satisfies
// all the requires conditions
if (N == 0) {
// Print Answer
for (int i = 0; i < count_Q; i++)
cout << Q;
for (int i = 0; i < count_P; i++)
cout << P;
}
else {
cout << "Not Possible";
}
}
// Driver Code
int main()
{
int N = 32;
int P = 7;
int Q = 4;
// Function Call
printMinInteger(P, Q, N);
return 0;
}
// Java program for the above approach
import java.io.*;
class GFG {
// Function to print the minimum
// integer having only digits P and
// Q and the sum of digits as N
static void printMinInteger(int P, int Q, int N)
{
// If Q is greater that P then
// swap the values of P and Q
if (Q > P) {
int temp;
temp = P;
P = Q;
Q = temp;
}
// If P and Q are both zero or
// if Q is zero and N is not
// divisible by P then there
// is no possible integer which
// satisfies the given conditions
if (Q == 0 && (P == 0 || N % P != 0)) {
System.out.println("Not Possible");
return;
}
int count_P = 0, count_Q = 0;
// Loop to find the maximum value
// of count_P that also satisfy
// P*count_P + Q*count_Q = N
while (N > 0) {
if (N % P == 0) {
count_P += N / P;
N = 0;
}
else {
N = N - Q;
count_Q++;
}
}
// If N is 0, there is a valid
// integer possible that satisfies
// all the requires conditions
if (N == 0) {
// Print Answer
for (int i = 0; i < count_Q; i++)
System.out.print(Q);
for (int i = 0; i < count_P; i++)
System.out.print(P);
}
else {
System.out.println("Not Possible");
}
}
// Driver code
public static void main(String[] args)
{
int N = 32;
int P = 7;
int Q = 4;
// Function Call
printMinInteger(P, Q, N);
}
}
// This code is contributed by code_hunt.
# Python3 program for the above approach
# Function to print minimum
# integer having only digits P and
# Q and the sum of digits as N
def printMinInteger(P, Q, N):
# If Q is greater that P then
# swap the values of P and Q
if (Q > P):
t = P
P = Q
Q = t
# If P and Q are both zero or
# if Q is zero and N is not
# divisible by P then there
# is no possible integer which
# satisfies the given conditions
if (Q == 0 and (P == 0 or N % P != 0)):
print("Not Possible")
return
count_P = 0
count_Q = 0
# Loop to find the maximum value
# of count_P that also satisfy
# P*count_P + Q*count_Q = N
while (N > 0):
if (N % P == 0):
count_P += N / P
N = 0
else:
N = N - Q
count_Q += 1
# If N is 0, there is a valid
# integer possible that satisfies
# all the requires conditions
if (N == 0):
# Print Answer
for i in range(count_Q):
print(Q, end = "")
for i in range(int(count_P)):
print(P, end = "")
else:
print("Not Possible")
# Driver Code
N = 32
P = 7
Q = 4
# Function Call
printMinInteger(P, Q, N)
# This code is contributed by code_hunt
// C# program for the above approach
using System;
public class GFG {
// Function to print the minimum
// integer having only digits P and
// Q and the sum of digits as N
static void printMinint(int P, int Q, int N)
{
// If Q is greater that P then
// swap the values of P and Q
if (Q > P) {
int temp;
temp = P;
P = Q;
Q = temp;
}
// If P and Q are both zero or
// if Q is zero and N is not
// divisible by P then there
// is no possible integer which
// satisfies the given conditions
if (Q == 0 && (P == 0 || N % P != 0)) {
Console.WriteLine("Not Possible");
return;
}
int count_P = 0, count_Q = 0;
// Loop to find the maximum value
// of count_P that also satisfy
// P*count_P + Q*count_Q = N
while (N > 0) {
if (N % P == 0) {
count_P += N / P;
N = 0;
}
else {
N = N - Q;
count_Q++;
}
}
// If N is 0, there is a valid
// integer possible that satisfies
// all the requires conditions
if (N == 0) {
// Print Answer
for (int i = 0; i < count_Q; i++)
Console.Write(Q);
for (int i = 0; i < count_P; i++)
Console.Write(P);
}
else {
Console.WriteLine("Not Possible");
}
}
// Driver code
public static void Main(String[] args)
{
int N = 32;
int P = 7;
int Q = 4;
// Function Call
printMinint(P, Q, N);
}
}
// This code contributed by shikhasingrajput
<script>
// Javascript Program of the above approach
// Function to print the minimum
// integer having only digits P and
// Q and the sum of digits as N
function printMinInteger(P, Q, N) {
// If Q is greater that P then
// swap the values of P and Q
if (Q > P) {
swap(P, Q);
}
// If P and Q are both zero or
// if Q is zero and N is not
// divisible by P then there
// is no possible integer which
// satisfies the given conditions
if (Q == 0 && (P == 0 || N % P != 0)) {
document.write("Not Possible");
return;
}
let count_P = 0,
count_Q = 0;
// Loop to find the maximum value
// of count_P that also satisfy
// P*count_P + Q*count_Q = N
while (N > 0) {
if (N % P == 0) {
count_P += Math.floor(N / P);
N = 0;
} else {
N = N - Q;
count_Q++;
}
}
// If N is 0, there is a valid
// integer possible that satisfies
// all the requires conditions
if (N == 0) {
// Print Answer
for (let i = 0; i < count_Q; i++) document.write(Q);
for (let i = 0; i < count_P; i++) document.write(P);
} else {
document.write("Not Possible");
}
}
// Driver Code
let N = 32;
let P = 7;
let Q = 4;
// Function Call
printMinInteger(P, Q, N);
// This code is contributed by gfgking.
</script>
Time Complexity: O(N)
Space Complexity: O(1)
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