Minimum K such that sum of array elements after division by K does not exceed S
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 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.
Efficient approach: An efficient approach is to find the value of K by performing 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 condition int 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 code int 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; } |
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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 |
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Python3
# Python3 implementation of the approach # Function to return the minimum value of k # that satisfies the given condition def 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 code a = [10, 7, 8, 10, 12, 19] n = len(a) s = 27 print(findMinimumK(a, n, s)) # This code is contributed by Mohit Kumar |
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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 |
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Output:
3
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