Length of smallest sequence having sum X and product Y
Given two integers X and Y, the task is to find the length of the smallest sequence consisting of positive integers having sum X and product Y. If no such sequence can be generated, print -1.
Examples:
Input: X = 5, Y = 5
Output: 1
Explanation:
The smallest possible sequence having sum 5 and product 5 is {5}. Therefore, the required answer is length of the sequence(= 1).Input: X = 9, Y = 8
Output: 2
Explanation:
The smallest possible sequence is {1, 8} with sum X (= 1 + 8 = 9) and product Y(= 1 * 8 = 8).
Approach: The idea is to implement Binary Search to solve the given problem on the required size in the range [1, floor(x/e)] where e is the Euler constant.
- Initialize two variables low and high to 1 and floor(x/e) respectively.
- If X is equal to Y, then the maximum size of the sequence can always be 1. Therefore, print it.
- Otherwise, iterate until the difference between high and low exceeds 1 and calculate mid each time.
- Reset the values of high and low as per the boundary conditions at each step and print the final value of high at the end.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;#define ll long long#define pb push_back// Function for checking valid or notdouble temp(int n, int x){ return pow(x * 1.0 / n, n);}// Function for checking boundary// of binary searchbool check(int n, int y, int x){ double v = temp(n, x); return (v >= y);}// Function to calculate the// minimum sequence size// using binary searchvoid find(int x, int y){ // Initialize high and low int high = (int)floor(x / exp(1.0)); int low = 1; // Base case if (x == y) cout << 1 << endl; // Print -1 if a sequence // cannot be generated else if (!check(high, y, x)) cout << -1 << endl; // Otherwise else { // Iterate until difference // between high and low exceeds 1 while (high - low > 1) { // Calculate mid int mid = (high + low) / 2; // Reset values of high // and low accordingly if (check(mid, y, x)) high = mid; else low = mid; } // Print the answer cout << high << endl; }}// Driver Codeint main(){ int x = 9, y = 8; // Function call find(x, y); return 0;} |
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Java
// Java program for the above approachimport java.util.*;class GFG{// Function for checking valid or notstatic double temp(int n, int x){ return Math.pow(x * 1.0 / n, n);}// Function for checking boundary// of binary searchstatic boolean check(int n, int y, int x){ double v = temp(n, x); return (v >= y);}// Function to calculate the// minimum sequence size// using binary searchstatic void find(int x, int y){ // Initialize high and low int high = (int)Math.floor(x / Math.exp(1.0)); int low = 1; // Base case if (x == y) System.out.print(1 + "\n"); // Print -1 if a sequence // cannot be generated else if (!check(high, y, x)) System.out.print(-1 + "\n"); // Otherwise else { // Iterate until difference // between high and low exceeds 1 while (high - low > 1) { // Calculate mid int mid = (high + low) / 2; // Reset values of high // and low accordingly if (check(mid, y, x)) high = mid; else low = mid; } // Print the answer System.out.print(high + "\n"); }}// Driver Codepublic static void main(String[] args){ int x = 9, y = 8; // Function call find(x, y);}}// This code is contributed by Amit Katiyar |
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Python3
# Python3 program for the above approachfrom math import floor, exp# Function for checking valid or notdef temp(n, x): return pow(x * 1 / n, n)# Function for checking boundary# of binary searchdef check(n, y, x): v = temp(n, x) return (v >= y)# Function to calculate the# minimum sequence size# using binary searchdef find(x, y): # Initialize high and low high = floor(x / exp(1.0)) low = 1 # Base case if (x == y): print(1) # Print -1 if a sequence # cannot be generated elif (not check(high, y, x)): print(-1) # Otherwise else: # Iterate until difference # between high and low exceeds 1 while (high - low > 1): # Calculate mid mid = (high + low) // 2 # Reset values of high # and low accordingly if (check(mid, y, x)): high = mid else: low = mid # Print the answer print(high)# Driver Codeif __name__ == '__main__': x = 9 y = 8 # Function call find(x, y)# This code is contributed by mohit kumar 29 |
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C#
// C# program for // the above approachusing System;class GFG{// Function for checking // valid or notstatic double temp(int n, int x){ return Math.Pow(x * 1.0 / n, n);}// Function for checking // boundary of binary searchstatic bool check(int n, int y, int x){ double v = temp(n, x); return (v >= y);}// Function to calculate the// minimum sequence size// using binary searchstatic void find(int x, int y){ // Initialize high and low int high = (int)Math.Floor(x / Math.Exp(1.0)); int low = 1; // Base case if (x == y) Console.Write(1 + "\n"); // Print -1 if a sequence // cannot be generated else if (!check(high, y, x)) Console.Write(-1 + "\n"); // Otherwise else { // Iterate until difference // between high and low exceeds 1 while (high - low > 1) { // Calculate mid int mid = (high + low) / 2; // Reset values of high // and low accordingly if (check(mid, y, x)) high = mid; else low = mid; } // Print the answer Console.Write(high + "\n"); }}// Driver Codepublic static void Main(String[] args){ int x = 9, y = 8; // Function call find(x, y);}}// This code is contributed by Rajput-Ji |
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Output
2
Time Complexity: O(log2(X/e))
Auxiliary Space: O(1)
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