Find maximum in an array without using Relational Operators

Given an array A[] of non-negative integers, find the maximum in the array without using Relational Operator.

Examples:

Input : A[] = {2, 3, 1, 4, 5}
Output : 5

Input : A[] = {23, 17, 93}
Output : 93

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We use repeated subtraction to find out the maximum. To find maximum between two numbers, we take a variable counter initialized to zero. We keep decreasing both the value till both of them becomes equal to zero (Note : The first value to become zero is no further decreased), increasing the counter simultaneously. While both the values becomes zero, the counter has increased to be the maximum of both of them. We first find the maximum of first two numbers and then compare it with the rest elements of the array one by one to find the overall maximum.

Below is the implementation of the above idea.

C++

 #include using namespace std;    // Function to find maximum between two non-negative // numbers without using relational operator. int maximum(int x, int y) {     int c = 0;        // Continues till both becomes zero.     while(x || y)     {         // decrement if the value is not already zero         if(x)         x--;            if(y)         y--;         c++;     }     return c; }    // Function to find maximum in an array. int arrayMaximum(int A[], int N) {     // calculating maximum of first two numbers     int mx = A;            // Iterating through each of the member of the array      // to calculate the maximum     for (int i = N-1; i; i--)            // Finding the maximum between current maximum         // and current value.         mx = maximum(mx, A[i]);            return mx; }    // Driver code int main()  {     // Array declaration      int A[] = {4, 8, 9, 18};     int N = sizeof(A) / sizeof(A);            // Calling Function to find the maximum of the Array     cout << arrayMaximum(A, N);     return 0; }

Java

 import java.io.*;    class GFG {            // Function to find maximum between two      // non-negative numbers without using      // relational operator.     static int maximum(int x, int y)     {         int c = 0;            // Continues till both becomes zero.         while (x > 0 || y > 0) {                            // decrement if the value is not              // already zero             if (x > 0)                 x--;                if (y > 0)                 y--;             c++;         }                    return c;     }        // Function to find maximum in an array.     static int arrayMaximum(int A[], int N)     {                    // calculating maximum of first          // two numbers         int mx = A;            // Iterating through each of the          // member of the array to calculate         // the maximum         for (int i = N - 1; i > 0; i--)                // Finding the maximum between              // current maximum and current              // value.             mx = maximum(mx, A[i]);            return mx;     }        // Driver code     public static void main(String[] args)     {                    // Array declaration         int A[] = { 4, 8, 9, 18 };         int N = A.length;            // Calling Function to find the maximum         // of the Array         System.out.print(arrayMaximum(A, N));     } }    // This code is contributed by vt_m.

Python3

 # Function to find maximum between two  # non-negative numbers without using  # relational operator. def maximum(x, y):     c = 0        # Continues till both becomes zero.     while(x or y):                    # decrement if the value is          # not already zero         if(x):             x -= 1            if(y):             y -= 1         c += 1        return c    # Function to find maximum in an array. def arrayMaximum(A, N):            # calculating maximum of      # first two numbers     mx = A            # Iterating through each of      # the member of the array      # to calculate the maximum     i = N - 1     while(i):                    # Finding the maximum between          # current maximum and current value.         mx = maximum(mx, A[i])         i -= 1            return mx    # Driver code if __name__ == '__main__':            # Array declaration      A = [4, 8, 9, 18]     N = len(A)            # Calling Function to find the      # maximum of the Array     print(arrayMaximum(A, N))        # This code is contributed by # Surendra_Gangwar

C#

 // C# program to Find maximum // in an array without using  // Relational Operators using System;    class GFG  {            // Function to find maximum      // between two non-negative     // numbers without using      // relational operator.     static int maximum(int x,                         int y)     {         int c = 0;            // Continues till          // both becomes zero.         while (x > 0 || y > 0)          {                            // decrement if              // the value is not              // already zero             if (x > 0)                 x--;                if (y > 0)                 y--;             c++;         }                    return c;     }        // Function to find      // maximum in an array.     static int arrayMaximum(int []A,                              int N)     {                    // calculating          // maximum of first          // two numbers         int mx = A;            // Iterating through          // each of the member         // of the array to          // calculate the maximum         for (int i = N - 1;                  i > 0; i--)                // Finding the maximum              // between current              // maximum and current              // value.             mx = maximum(mx, A[i]);            return mx;     }        // Driver code     public static void Main()     {                    // Array declaration         int []A = { 4, 8, 9, 18 };         int N = A.Length;            // Calling Function to         // find the maximum         // of the Array         Console.WriteLine(arrayMaximum(A, N));     } }    // This code is contributed // by anuj_67.

PHP



Output:

18

The time complexity of the code will be O(N*max) where max is the maximum of the array elements.

Limitations : This will only work if the array contains all non negative integers.

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Improved By : vt_m, SURENDRA_GANGWAR

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