Given an array arr[] of N integers and three integers X, Y and Z. The task is to find the maximum value of (arr[i] * X) + (arr[j] * Y) + (arr[k] * Z) where 0 ? i ? j ? k ? N – 1.
Examples:
Input: arr[] = {1, 5, -3, 4, -2}, X = 2, Y = 1, Z = -1
Output: 18
(2 * 5) + (1 * 5) + (-1 * -3) = 18
is the maximum possible sum.Input: arr[] = {2, 4, -9, -64, 7, 3}, X = -1, Y = 1, Z = 1
Output: 78
Approach: Find the maximum and the minimum negative and positive values from the array. Also, check whether 0 is present in the array or not. Now, for the given values of X, Y and Z. Choose the values found previously from the array which maximizes the overall sum.
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 maximum possible// value of the given equationint maxSum(int arr[], int n, int x, int y, int z){ // To store the minimum and the maximum negative // and positive values from the array int minNeg = INT_MAX, maxNeg = INT_MAX; int minPos = INT_MIN, maxPos = INT_MIN; // To store whether 0 is present in the array bool isZeroPresent = false; // Update the values of the // above defined variables for (int i = 0; i < n; i++) { if (arr[i] == 0) { isZeroPresent = true; } else if (arr[i] < 0) { minNeg = min(minNeg, arr[i]); maxNeg = max(maxNeg, arr[i]); } else { minPos = min(minPos, arr[i]); maxPos = max(maxPos, arr[i]); } } // To store the resultant sum int sum = 0; // x will not contribute to the // sum if it is equal to 0 if (x != 0) { // If x is negative if (x < 0) { // Either multiply it with the minimum // negative number from the array if (minNeg != INT_MAX) sum += (x * minNeg); // Or multiply it with the minimum // positive element if zero is // not present in the array else if (!isZeroPresent) sum += (x * minPos); } // If x is positive else { // Multiply it with the maximum // positive value from the array if (maxPos != INT_MIN) sum += (x * maxPos); // Or multiply it with the maximum // negative element if zero is // not present in the array else if (!isZeroPresent) sum += (x * maxPos); } } // Same as x if (y != 0) { if (y < 0) { if (minNeg != INT_MAX) sum += (y * minNeg); else if (!isZeroPresent) sum += (y * minPos); } else { if (maxPos != INT_MIN) sum += (y * maxPos); else if (!isZeroPresent) sum += (y * maxPos); } } // Same as x if (z != 0) { if (z < 0) { if (minNeg != INT_MAX) sum += (z * minNeg); else if (!isZeroPresent) sum += (z * minPos); } else { if (maxPos != INT_MIN) sum += (z * maxPos); else if (!isZeroPresent) sum += (z * maxPos); } } return sum;}// Driver codeint main(){ int arr[] = { 2, 4, -9, -64, 7, 3 }; int n = sizeof(arr) / sizeof(arr[0]); int x = -1, y = 1, z = 1; cout << maxSum(arr, n, x, y, z); return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to return the maximum possible// value of the given equationstatic int maxSum(int arr[], int n, int x, int y, int z){ // To store the minimum and the maximum negative // and positive values from the array int minNeg = Integer.MAX_VALUE, maxNeg = Integer.MAX_VALUE; int minPos = Integer.MIN_VALUE, maxPos = Integer.MIN_VALUE; // To store whether 0 is present in the array boolean isZeroPresent = false; // Update the values of the // above defined variables for (int i = 0; i < n; i++) { if (arr[i] == 0) { isZeroPresent = true; } else if (arr[i] < 0) { minNeg = Math.min(minNeg, arr[i]); maxNeg = Math.max(maxNeg, arr[i]); } else { minPos = Math.min(minPos, arr[i]); maxPos = Math.max(maxPos, arr[i]); } } // To store the resultant sum int sum = 0; // x will not contribute to the // sum if it is equal to 0 if (x != 0) { // If x is negative if (x < 0) { // Either multiply it with the minimum // negative number from the array if (minNeg != Integer.MAX_VALUE) sum += (x * minNeg); // Or multiply it with the minimum // positive element if zero is // not present in the array else if (!isZeroPresent) sum += (x * minPos); } // If x is positive else { // Multiply it with the maximum // positive value from the array if (maxPos != Integer.MIN_VALUE) sum += (x * maxPos); // Or multiply it with the maximum // negative element if zero is // not present in the array else if (!isZeroPresent) sum += (x * maxPos); } } // Same as x if (y != 0) { if (y < 0) { if (minNeg != Integer.MAX_VALUE) sum += (y * minNeg); else if (!isZeroPresent) sum += (y * minPos); } else { if (maxPos != Integer.MIN_VALUE) sum += (y * maxPos); else if (!isZeroPresent) sum += (y * maxPos); } } // Same as x if (z != 0) { if (z < 0) { if (minNeg != Integer.MAX_VALUE) sum += (z * minNeg); else if (!isZeroPresent) sum += (z * minPos); } else { if (maxPos != Integer.MIN_VALUE) sum += (z * maxPos); else if (!isZeroPresent) sum += (z * maxPos); } } return sum;}// Driver codepublic static void main(String[] args){ int arr[] = { 2, 4, -9, -64, 7, 3 }; int n = arr.length; int x = -1, y = 1, z = 1; System.out.print(maxSum(arr, n, x, y, z));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach# Function to return the maximum possible# value of the given equationdef maxSum(arr, n, x, y, z): # To store the minimum and the maximum negative # and positive values from the array minNeg = 10**9 maxNeg = 10**9 minPos = -10**9 maxPos = -10**9 # To store whether 0 is present in the array isZeroPresent = False # Update the values of the # above defined variables for i in range(n): if (arr[i] == 0): isZeroPresent = True elif (arr[i] < 0): minNeg = min(minNeg, arr[i]) maxNeg = max(maxNeg, arr[i]) else: minPos = min(minPos, arr[i]) maxPos = max(maxPos, arr[i]) # To store the resultant summ summ = 0 # x will not contribute to the # summ if it is equal to 0 if (x != 0): # If x is negative if (x < 0): # Either multiply it with the minimum # negative number from the array if (minNeg != 10**9): summ += (x * minNeg) # Or multiply it with the minimum # positive element if zero is # not present in the array elif (isZeroPresent == False): summ += (x * minPos) # If x is positive else: # Multiply it with the maximum # positive value from the array if (maxPos != -10**9): summ += (x * maxPos) # Or multiply it with the maximum # negative element if zero is # not present in the array elif (isZeroPresent == False): summ += (x * maxPos) # Same as x if (y != 0): if (y < 0): if (minNeg != 10**9): summ += (y * minNeg) elif (isZeroPresent == False): summ += (y * minPos) else: if (maxPos != -10**9): summ += (y * maxPos) elif (isZeroPresent == False): summ += (y * maxPos) # Same as x if (z != 0): if (z < 0): if (minNeg != 10**9): summ += (z * minNeg) elif (isZeroPresent == False): summ += (z * minPos) else: if (maxPos != -10**9): summ += (z * maxPos) elif (isZeroPresent == False): summ += (z * maxPos) return summ# Driver codearr = [2, 4, -9, -64, 7, 3]n = len(arr)x = -1y = 1z = 1print(maxSum(arr, n, x, y, z))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the maximum possible // value of the given equation static int maxSum(int []arr, int n, int x, int y, int z) { int INT_MAX = int.MaxValue; int INT_MIN = int.MinValue; // To store the minimum and the maximum negative // and positive values from the array int minNeg = INT_MAX, maxNeg = INT_MAX; int minPos = INT_MIN, maxPos = INT_MIN; // To store whether 0 is present in the array bool isZeroPresent = false; // Update the values of the // above defined variables for (int i = 0; i < n; i++) { if (arr[i] == 0) { isZeroPresent = true; } else if (arr[i] < 0) { minNeg = Math.Min(minNeg, arr[i]); maxNeg = Math.Max(maxNeg, arr[i]); } else { minPos = Math.Min(minPos, arr[i]); maxPos = Math.Max(maxPos, arr[i]); } } // To store the resultant sum int sum = 0; // x will not contribute to the // sum if it is equal to 0 if (x != 0) { // If x is negative if (x < 0) { // Either multiply it with the minimum // negative number from the array if (minNeg != INT_MAX) sum += (x * minNeg); // Or multiply it with the minimum // positive element if zero is // not present in the array else if (!isZeroPresent) sum += (x * minPos); } // If x is positive else { // Multiply it with the maximum // positive value from the array if (maxPos != INT_MIN) sum += (x * maxPos); // Or multiply it with the maximum // negative element if zero is // not present in the array else if (!isZeroPresent) sum += (x * maxPos); } } // Same as x if (y != 0) { if (y < 0) { if (minNeg != INT_MAX) sum += (y * minNeg); else if (!isZeroPresent) sum += (y * minPos); } else { if (maxPos != INT_MIN) sum += (y * maxPos); else if (!isZeroPresent) sum += (y * maxPos); } } // Same as x if (z != 0) { if (z < 0) { if (minNeg != INT_MAX) sum += (z * minNeg); else if (!isZeroPresent) sum += (z * minPos); } else { if (maxPos != INT_MIN) sum += (z * maxPos); else if (!isZeroPresent) sum += (z * maxPos); } } return sum; } // Driver code static public void Main () { int []arr = { 2, 4, -9, -64, 7, 3 }; int n = arr.Length; int x = -1, y = 1, z = 1; Console.Write(maxSum(arr, n, x, y, z)); } }// This code is contributed by AnkitRai01 |
Javascript
<script>// JavaScript implementation of the approach// Function to return the maximum possible// value of the given equationfunction maxSum(arr, n, x, y, z) { // To store the minimum and the maximum negative // and positive values from the array let minNeg = Number.MAX_SAFE_INTEGER, maxNeg = Number.MAX_SAFE_INTEGER; let minPos = Number.MIN_SAFE_INTEGER, maxPos = Number.MIN_SAFE_INTEGER; // To store whether 0 is present in the array let isZeroPresent = false; // Update the values of the // above defined variables for (let i = 0; i < n; i++) { if (arr[i] == 0) { isZeroPresent = true; } else if (arr[i] < 0) { minNeg = Math.min(minNeg, arr[i]); maxNeg = Math.max(maxNeg, arr[i]); } else { minPos = Math.min(minPos, arr[i]); maxPos = Math.max(maxPos, arr[i]); } } // To store the resultant sum let sum = 0; // x will not contribute to the // sum if it is equal to 0 if (x != 0) { // If x is negative if (x < 0) { // Either multiply it with the minimum // negative number from the array if (minNeg != Number.MAX_SAFE_INTEGER) sum += (x * minNeg); // Or multiply it with the minimum // positive element if zero is // not present in the array else if (!isZeroPresent) sum += (x * minPos); } // If x is positive else { // Multiply it with the maximum // positive value from the array if (maxPos != Number.MIN_SAFE_INTEGER) sum += (x * maxPos); // Or multiply it with the maximum // negative element if zero is // not present in the array else if (!isZeroPresent) sum += (x * maxPos); } } // Same as x if (y != 0) { if (y < 0) { if (minNeg != Number.MAX_SAFE_INTEGER) sum += (y * minNeg); else if (!isZeroPresent) sum += (y * minPos); } else { if (maxPos != Number.MIN_SAFE_INTEGER) sum += (y * maxPos); else if (!isZeroPresent) sum += (y * maxPos); } } // Same as x if (z != 0) { if (z < 0) { if (minNeg != Number.MAX_SAFE_INTEGER) sum += (z * minNeg); else if (!isZeroPresent) sum += (z * minPos); } else { if (maxPos != Number.MIN_SAFE_INTEGER) sum += (z * maxPos); else if (!isZeroPresent) sum += (z * maxPos); } } return sum;}// Driver codelet arr = [2, 4, -9, -64, 7, 3];let n = arr.length;let x = -1, y = 1, z = 1;document.write(maxSum(arr, n, x, y, z));</script> |
Output:
78
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