Open In App

# 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```

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[0];``    ` `    ``// 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[0]);``    ` `    ``// 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[``0``];` `        ``// 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[``0``]``    ` `    ``# 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[0];` `        ``// 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

 ``

## Javascript

 ``

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.