Given an array arr[] of N elements and an integer S. The task is to find the minimum number K such that the sum of the array elements does not exceed S after dividing all the elements by K.
Note: Consider integer division.
Examples:
Input: arr[] = {10, 7, 8, 10, 12, 19}, S = 27
Output: 3
After dividing by 3, the array becomes
{3, 2, 2, 3, 4, 6} and the new sum is 20.
Input: arr[] = {19, 17, 11, 10}, S = 40
Output: 2
Naive approach: Iterate for all values of K from 1 to the maximum element in the array plus one not maximum element because if we put k as maximum element then the sum will be one and what if the S is given to us as zero. So we iterate from k=1 to max element in the array plus one. and then sum up the array elements by dividing with K, if the sum does not exceed S then the current value will be the answer. The time complexity of this approach will be O(M * N) where M is the maximum element in the array.
C++
#include <iostream>#include <numeric> // Required for std::accumulate// Define the FindK function to find the minimum value of Kint FindK(int arr[], int n, int S){ // Initialize K to 1 int K = 1; // Enter into a loop to find the minimum value of K while (true) { // Calculate the sum of the array after dividing each element by K int sumArr = std::accumulate(arr, arr+n, 0, [=](int a, int b) { return a + b/K; }); // If the resulting sum is less than or equal to S, return K as the answer if (sumArr <= S) { return K; } // Otherwise, increment the value of K and repeat the loop K++; }}int main(){ // Define the input array and integer int arr[] = {10, 7, 8, 10, 12, 19}; int n = sizeof(arr)/sizeof(arr[0]); // Find the size of the array int S = 27; // Call the FindK function with the input array and integer, and print the result std::cout << FindK(arr, n, S) << std::endl; // Output: 3 return 0;} |
Java
import java.util.*;public class Main { public static int FindK(int arr[], int n, int S) { int K = 1; while (true) { // declare a final variable k and assign the value of K to it final int k = K; int sumArr = Arrays.stream(arr).reduce( 0, // use the final variable k in the lambda expression (a, b) -> a + b/k); if (sumArr <= S) { return K; } K++; } } public static void main(String[] args) { int arr[] = {10, 7, 8, 10, 12, 19}; int n = arr.length; int S = 27; System.out.println(FindK(arr, n, S)); // Output: 3 }} |
Python3
# Define the input array and integerarr = [10, 7, 8, 10, 12, 19]S = 27# Define a function to find the minimum value of Kdef find_k(arr, S): # Initialize K to 1 K = 1 # Enter into a loop to find the minimum value of K while True: # Calculate the sum of the array after dividing each element by K sum_arr = sum([i//K for i in arr]) # If the resulting sum is less than or equal to S, return K as the answer if sum_arr <= S: return K # Otherwise, increment the value of K and repeat the loop K += 1# Call the find_k function with the input array and integer, and print the resultprint(find_k(arr, S)) # Output: 3 |
Javascript
// Define the input array and integerconst arr = [10, 7, 8, 10, 12, 19];const S = 27;// Define a function to find the minimum value of Kfunction find_k(arr, S) { // Initialize K to 1 let K = 1; // Enter into a loop to find the minimum value of K while (true) { // Calculate the sum of the array after dividing each element by K let sum_arr = arr.reduce((acc, curr) => acc + Math.floor(curr / K), 0); // If the resulting sum is less than or equal to S, return K as the answer if (sum_arr <= S) { return K; } // Otherwise, increment the value of K and repeat the loop K++; }}// Call the find_k function with the input array and integer, and print the resultconsole.log(find_k(arr, S)); // Output: 3 |
C#
using System;using System.Linq;public class MinimumKValue{ public static void Main() { // Define the input array and integer int[] arr = {10, 7, 8, 10, 12, 19}; int S = 27; // Call the FindK function with the input array and integer, and print the result Console.WriteLine(FindK(arr, S)); // Output: 3 } // Define a function to find the minimum value of K public static int FindK(int[] arr, int S) { // Initialize K to 1 int K = 1; // Enter into a loop to find the minimum value of K while (true) { // Calculate the sum of the array after dividing each element by K int sumArr = arr.Sum(i => i / K); // If the resulting sum is less than or equal to S, return K as the answer if (sumArr <= S) { return K; } // Otherwise, increment the value of K and repeat the loop K++; } }} |
Output
3
Efficient approach: An efficient approach is to find the value of K by performing a binary search on the answer. Initiate a binary search on the value of K and a check is done inside it to see if the sum exceeds K then the binary search is performed on the second half or the first half accordingly.
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 minimum value of k// that satisfies the given conditionint findMinimumK(int a[], int n, int s){ // Find the maximum element int maximum = a[0]; for (int i = 0; i < n; i++) { maximum = max(maximum, a[i]); } // Lowest answer can be 1 and the // highest answer can be (maximum + 1) int low = 1, high = maximum + 1; int ans = high; // Binary search while (low <= high) { // Get the mid element int mid = (low + high) / 2; int sum = 0; // Calculate the sum after dividing // the array by new K which is mid for (int i = 0; i < n; i++) { sum += (int)(a[i] / mid); } // Search in the second half if (sum > s) low = mid + 1; // First half else { ans = min(ans, mid); high = mid - 1; } } return ans;}// Driver codeint main(){ int a[] = { 10, 7, 8, 10, 12, 19 }; int n = sizeof(a) / sizeof(a[0]); int s = 27; cout << findMinimumK(a, n, s); return 0;} |
Java
// Java implementation of the approach class GFG { // Function to return the minimum value of k // that satisfies the given condition static int findMinimumK(int a[], int n, int s) { // Find the maximum element int maximum = a[0]; for (int i = 0; i < n; i++) { maximum = Math.max(maximum, a[i]); } // Lowest answer can be 1 and the // highest answer can be (maximum + 1) int low = 1, high = maximum + 1; int ans = high; // Binary search while (low <= high) { // Get the mid element int mid = (low + high) / 2; int sum = 0; // Calculate the sum after dividing // the array by new K which is mid for (int i = 0; i < n; i++) { sum += (int)(a[i] / mid); } // Search in the second half if (sum > s) low = mid + 1; // First half else { ans = Math.min(ans, mid); high = mid - 1; } } return ans; } // Driver code public static void main (String[] args) { int a[] = { 10, 7, 8, 10, 12, 19 }; int n = a.length; int s = 27; System.out.println(findMinimumK(a, n, s)); } } // This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach# Function to return the minimum value of k# that satisfies the given conditiondef findMinimumK(a, n, s): # Find the maximum element maximum = a[0] for i in range(n): maximum = max(maximum, a[i]) # Lowest answer can be 1 and the # highest answer can be (maximum + 1) low = 1 high = maximum + 1 ans = high # Binary search while (low <= high): # Get the mid element mid = (low + high) // 2 sum = 0 # Calculate the sum after dividing # the array by new K which is mid for i in range(n): sum += (a[i] // mid) # Search in the second half if (sum > s): low = mid + 1 # First half else: ans = min(ans, mid) high = mid - 1 return ans# Driver codea = [10, 7, 8, 10, 12, 19]n = len(a)s = 27print(findMinimumK(a, n, s))# This code is contributed by Mohit Kumar |
Javascript
<script>// javascript implementation of the approach // Function to return the minimum value of k // that satisfies the given condition function findMinimumK(a , n , s) { // Find the maximum element var maximum = a[0]; for (i = 0; i < n; i++) { maximum = Math.max(maximum, a[i]); } // Lowest answer can be 1 and the // highest answer can be (maximum + 1) var low = 1, high = maximum + 1; var ans = high; // Binary search while (low <= high) { // Get the mid element var mid = parseInt((low + high) / 2); var sum = 0; // Calculate the sum after dividing // the array by new K which is mid for (i = 0; i < n; i++) { sum += parseInt( (a[i] / mid)); } // Search in the second half if (sum > s) low = mid + 1; // First half else { ans = Math.min(ans, mid); high = mid - 1; } } return ans; } // Driver code var a = [ 10, 7, 8, 10, 12, 19 ]; var n = a.length; var s = 27; document.write(findMinimumK(a, n, s));// This code is contributed by todaysgaurav</script> |
C#
// C# implementation of the approach using System;class GFG { // Function to return the minimum value of k // that satisfies the given condition static int findMinimumK(int []a, int n, int s) { // Find the maximum element int maximum = a[0]; for (int i = 0; i < n; i++) { maximum = Math.Max(maximum, a[i]); } // Lowest answer can be 1 and the // highest answer can be (maximum + 1) int low = 1, high = maximum + 1; int ans = high; // Binary search while (low <= high) { // Get the mid element int mid = (low + high) / 2; int sum = 0; // Calculate the sum after dividing // the array by new K which is mid for (int i = 0; i < n; i++) { sum += (int)(a[i] / mid); } // Search in the second half if (sum > s) low = mid + 1; // First half else { ans = Math.Min(ans, mid); high = mid - 1; } } return ans; } // Driver code public static void Main () { int []a = { 10, 7, 8, 10, 12, 19 }; int n = a.Length; int s = 27; Console.WriteLine(findMinimumK(a, n, s)); } } // This code is contributed by AnkitRai01 |
Output:
3
Time Complexity: O(N*(log N)), N=Array length
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
… [Trackback]
[…] Read More here on that Topic: geeksforgeeks.org/minimum-k-such-that-sum-of-array-elements-after-division-by-k-does-not-exceed-s/ […]
… [Trackback]
[…] Here you will find 11062 more Info on that Topic: geeksforgeeks.org/minimum-k-such-that-sum-of-array-elements-after-division-by-k-does-not-exceed-s/ […]