Given a number N and a digit K (1 < K <= 9), the task is to find two integers A and B such that A + B = N and digit K is not present in either A or B.
Examples:
Input: N = 443, K = 4
Output: 256, 187
Explanation:
Sum = 256 + 187 = 443 = N
Also, K(4) is not present in 256 or 187
Hence, 256 and 187 are valid output.Input: N = 678, K = 9
Output: 311, 367
Explanation:
Sum = 311 + 367 = 678 = N
Also, K(9) is not present in 311 or 367
Hence, 311 and 367 are valid output.
Naive Approach:
A simple approach is to run a loop and consider each element (say A) from 1 to N-1. Since the sum of A and B has to be N, corresponding B will be (N – A). Now check if both A and B contain K or not, if no, then print A and B. This approach will fail for larger inputs like N = 1018.
Time complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach:
The given problem can be solved much more efficiently. The idea is to break each digit D of N into two parts D1 and D2 such that D1 + D2 = D. Take two strings A and B, A will contain all D1’s, and string B will contain all D2’s.
For example, N = 4467 and K = 4
D D1 D2 4 2 2 4 2 2 6 6 0 7 7 0
Here, A will be 2267 and B will be 2200.
Detailed steps of this approach are as follows:
1. Traverse N and consider each of its digit D.
2. If digit D is same as K, break it into D1 = D / 2 and D2 = D / 2 + D % 2.
D % 2 is added to D2 to ensure that D1 + D2 = D for both even and odd D.
3. Else, break it into D1 = D and D2 = 0.
4. Keep appending D1’s to string A and D2’s to string B.
5. Finally, print string A and B, after removing leading zeros(if any).
Below is the implementation of the above approach:
C++
// C++ implementation of// the above approach#include <iostream>using namespace std;string removeLeadingZeros(string str){ // Count leading zeros int i = 0; int n = str.length(); while (str[i] == '0' && i < n) i++; // It removes i characters // starting from index 0 str.erase(0, i); return str;}void findPairs(int sum, int K){ string A, B; A = ""; B = ""; string N = to_string(sum); int n = N.length(); // Check each digit of the N for (int i = 0; i < n; i++) { int D = N[i] - '0'; // If digit is K break it if (D == K) { int D1, D2; D1 = D / 2; // For odd numbers D2 = D / 2 + D % 2; // Add D1 to A and D2 to B A = A + char(D1 + '0'); B = B + char(D2 + '0'); } // If the digit is not K, // no need to break // string D in A and 0 in B else { A = A + char(D + '0'); B = B + '0'; } } // Remove leading zeros A = removeLeadingZeros(A); B = removeLeadingZeros(B); // Print the answer cout << A << ", " << B << endl;}// Driver codeint main(){ int N = 33673; int K = 3; findPairs(N, K); return 0;} |
Java
// Java implementation of the above approachclass GFG{static String removeLeadingZeros(String str){ // Count leading zeros int i = 0; int n = str.length(); while (str.charAt(i) == '0' && i < n) i++; // It removes i characters // starting from index 0 str = str.substring(i); return str;}static void findPairs(int sum, int K){ String A, B; A = ""; B = ""; String N = String.valueOf(sum); int n = N.length(); // Check each digit of the N for(int i = 0; i < n; i++) { int D = N.charAt(i) - '0'; // If digit is K break it if (D == K) { int D1, D2; D1 = D / 2; // For odd numbers D2 = D / 2 + D % 2; // Add D1 to A and D2 to B A = A + (char)(D1 + '0'); B = B + (char)(D2 + '0'); } // If the digit is not K, // no need to break // String D in A and 0 in B else { A = A + (char)(D + '0'); B = B + '0'; } } // Remove leading zeros A = removeLeadingZeros(A); B = removeLeadingZeros(B); // Print the answer System.out.print(A + ", " + B + "\n");}// Driver codepublic static void main(String[] args){ int N = 33673; int K = 3; findPairs(N, K);}}// This code is contributed by sapnasingh4991 |
Python3
# Python3 implementation of the# above approachdef removeLeadingZeros(Str): # Count leading zeros i = 0 n = len(Str) while (Str[i] == '0' and i < n): i += 1 # It removes i characters # starting from index 0 Str = Str[i:] return Str def findPairs(Sum, K): A = "" B = "" N = str(Sum) n = len(N) # Check each digit of the N for i in range(n): D = int(N[i]) - int('0') # If digit is K break it if (D == K): D1 = D // 2; # For odd numbers D2 = (D // 2) + (D % 2) # Add D1 to A and D2 to B A = A + (str)(D1 + int('0')) B = B + (str)(D2 + int('0')) # If the digit is not K, # no need to break # String D in A and 0 in B else: A = A + (str)(D + int('0')) B = B + '0' # Remove leading zeros A = removeLeadingZeros(A) B = removeLeadingZeros(B) # Print the answer print(A + ", " + B) # Driver codeN = 33673K = 3findPairs(N, K)# This code is contributed divyeshrabadiya07 |
C#
// C# implementation of the above approachusing System;class GFG{ static String removeLeadingZeros(String str){ // Count leading zeros int i = 0; int n = str.Length; while (str[i] == '0' && i < n) i++; // It removes i characters // starting from index 0 str = str.Substring(i); return str;} static void findPairs(int sum, int K){ String A, B; A = ""; B = ""; String N = String.Join("", sum); int n = N.Length; // Check each digit of the N for(int i = 0; i < n; i++) { int D = N[i] - '0'; // If digit is K break it if (D == K) { int D1, D2; D1 = D / 2; // For odd numbers D2 = D / 2 + D % 2; // Add D1 to A and D2 to B A = A + (char)(D1 + '0'); B = B + (char)(D2 + '0'); } // If the digit is not K, // no need to break // String D in A and 0 in B else { A = A + (char)(D + '0'); B = B + '0'; } } // Remove leading zeros A = removeLeadingZeros(A); B = removeLeadingZeros(B); // Print the answer Console.Write(A + ", " + B + "\n");} // Driver codepublic static void Main(String[] args){ int N = 33673; int K = 3; findPairs(N, K);}}// This code is contributed by sapnasingh4991 |
11671, 22002
Time complexity: O(M)
Auxiliary Space: O(M)
Here, M is the number of digits in N.
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