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
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!
