Minimum operations to make all elements equal using the second array
Given two arrays A[] and B[] both having N elements. Find the minimum number of operations to make all elements of A equal to the second array B. An operation comprises of:
A[i] = A[i] - B[i], 0 <= i <= n-1
Note: If it’s not possible to make array elements equal print -1.
Examples:
Input: A[] = {5, 7, 10, 5, 15}, B[] = {2, 2, 1, 3, 5}
Output: 8
Explanation:
Following are the operations:
1) Choosing index 1 -> 5 5 10 5 15
2) Choosing index 2 -> 5 5 9 5 15
3) Choosing index 2 -> 5 5 8 5 15
4) Choosing index 2 -> 5 5 7 5 15
5) Choosing index 2 -> 5 5 6 5 15
6) Choosing index 2 -> 5 5 5 5 15
7) Choosing index 4 -> 5 5 5 5 10
8) Choosing index 4 -> 5 5 5 5 5Input: A[] = {3, 5, 8, 2}, B[] = {1, 2, 1, 1}
Output: 12
Approach:
- It can be observed that the maximum possible value if all elements of A can be made equal is the minimum element of A.
- So we can iterate over the final value x, when all elements of A become equal. Using above observation – 0 <= x <= min(A[i]). For each x, traverse A and check whether A[i] can be made equal to x using B[i] :
A[i] - k*B[i] = x Take mod with B[i] on both sides A[i] %B[i] = x %B[i] -> must be satisfied for A[i] to be converted to x
If condition is satisfied, then number of operations required for this element = (A[i] – x)/B[i]
Below is the implementation of the above approach:
C++
// C++ implementation to find the // minimum operations make all elements // equal using the second array #include <bits/stdc++.h> using namespace std; // Function to find the minimum // operations required to make // all elements of the array equal int minOperations(int a[], int b[], int n) { // Minimum element of A[] int minA = *min_element(a, a + n); // Traverse through all final values for (int x = minA; x >= 0; x--) { // Variable indicating // whether all elements // can be converted to x or not bool check = 1; // Total operations int operations = 0; // Traverse through // all array elements for (int i = 0; i < n; i++) { if (x % b[i] == a[i] % b[i]) { operations += (a[i] - x) / b[i]; } // All elements can't // be converted to x else { check = 0; break; } } if (check) return operations; } return -1; } // Driver Code int main() { int N = 5; int A[N] = { 5, 7, 10, 5, 15 }; int B[N] = { 2, 2, 1, 3, 5 }; cout << minOperations(A, B, N); return 0; } |
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Java
// Java implementation to find the // minimum operations make all elements // equal using the second array import java.util.*; class GFG{ // Function to find the minimum // operations required to make // all elements of the array equal static int minOperations(int a[], int b[], int n) { // Minimum element of A[] int minA = Arrays.stream(a).min().getAsInt(); // Traverse through all final values for (int x = minA; x >= 0; x--) { // Variable indicating // whether all elements // can be converted to x or not boolean check = true; // Total operations int operations = 0; // Traverse through // all array elements for (int i = 0; i < n; i++) { if (x % b[i] == a[i] % b[i]) { operations += (a[i] - x) / b[i]; } // All elements can't // be converted to x else { check = false; break; } } if (check) return operations; } return -1; } // Driver Code public static void main(String[] args) { int N = 5; int A[] = { 5, 7, 10, 5, 15 }; int B[] = { 2, 2, 1, 3, 5 }; System.out.print(minOperations(A, B, N)); } } // This code is contributed by AbhiThakur |
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Python3
# Python3 implementation to find the # minimum operations make all elements # equal using the second array # Function to find the minimum # operations required to make # all elements of the array equal def minOperations(a, b, n): # Minimum element of A minA = min(a); # Traverse through all final values for x in range(minA, -1, -1): # Variable indicating # whether all elements # can be converted to x or not check = True; # Total operations operations = 0; # Traverse through # all array elements for i in range(n): if (x % b[i] == a[i] % b[i]): operations += (a[i] - x) / b[i]; # All elements can't # be converted to x else: check = False; break; if (check): return operations; return -1; # Driver Code if __name__ == '__main__': N = 5; A = [ 5, 7, 10, 5, 15 ]; B = [ 2, 2, 1, 3, 5 ]; print(int(minOperations(A, B, N))); # This code is contributed by amal kumar choubey |
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C#
// C# implementation to find the // minimum operations make all elements // equal using the second array using System; using System.Linq; class GFG{ // Function to find the minimum // operations required to make // all elements of the array equal static int minOperations(int []a, int []b, int n) { // Minimum element of A[] int minA = a.Max(); // Traverse through all final values for(int x = minA; x >= 0; x--) { // Variable indicating // whether all elements // can be converted to x or not bool check = true; // Total operations int operations = 0; // Traverse through // all array elements for(int i = 0; i < n; i++) { if (x % b[i] == a[i] % b[i]) { operations += (a[i] - x) / b[i]; } // All elements can't // be converted to x else { check = false; break; } } if (check) return operations; } return -1; } // Driver Code public static void Main(string[] args) { int N = 5; int []A = { 5, 7, 10, 5, 15 }; int []B = { 2, 2, 1, 3, 5 }; Console.WriteLine(minOperations(A, B, N)); } } // This code is contributed by SoumikMondal |
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Output:
8
Time Complexity: O(N * min(Ai))
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