Find minimum in an array without using Relational Operators
Last Updated :
25 Jul, 2022
Given an array A[] of non-negative integers, find the minimum in the array without using Relational Operators.
Examples:
Input : A[] = {2, 3, 1, 4, 5}
Output : 1
Input : A[] = {23, 17, 93}
Output : 17
We use repeated subtraction to find out the minimum. To find minimum between two numbers, we take a variable counter initialized to zero. We keep decreasing the both the value till any one of them becomes equal to zero, increasing the counter simultaneously. The minimum value reaches zero first and the counter has increased to be the minimum of both of them. We first find the minimum of first two numbers and then compare it with the rest elements of the array one by one to find the overall minimum.
Below is the implementation of the above idea.
C++
// C++ program to find minimum in an
// array without using Relational Operators
#include <bits/stdc++.h>
using namespace std;
// Function to find minimum between two non-negative
// numbers without using relational operator.
int minimum(int x, int y)
{
int c = 0;
// Continues till any element becomes zero.
while (x && y)
{
x--;
y--;
c++;
}
return c;
}
// Function to find minimum in an array.
int arrayMinimum(int A[], int N)
{
// calculating minimum of first two numbers
int mn = A[0];
// Iterating through each of the member of
// the array to calculate the minimum
for (int i = N-1; i; i--)
// Finding the minimum between current
// minimum and current value.
mn = minimum(mn, A[i]);
return mn;
}
// Driver code
int main()
{
int A[] = { 2, 3, 1, 4 };
int N = sizeof(A) / sizeof(A[0]);
cout << arrayMinimum(A, N);
return 0;
}
// Java program to find minimum in an
// array without using Relational Operators
class GFG {
// Function to find minimum between two
// non-negative numbers without
// using relational operator.
static int minimum(int x, int y)
{
int c = 0;
// Continues till any element becomes zero.
while (x > 0 && y > 0) {
x--;
y--;
c++;
}
return c;
}
// Function to find minimum in an array.
static int arrayMinimum(int A[], int N) {
// calculating minimum of first two numbers
int mn = A[0];
// Iterating through each of the member of
// the array to calculate the minimum
for (int i = N - 1; i > 0; i--)
// Finding the minimum between current
// minimum and current value.
mn = minimum(mn, A[i]);
return mn;
}
// Driver code
public static void main(String arg[])
{
int A[] = {2, 3, 1, 4};
int N = A.length;
System.out.print(arrayMinimum(A, N));
}
}
// This code is contributed by Anant Agarwal.
# Function to find minimum
# between two non-negative
# numbers without using
# relational operator.
def minimum(x,y):
c = 0
# Continues till any
# element becomes zero.
while (x>0 and y>0):
x=x-1
y=y-1
c=c+1
return c
# Function to find
# minimum in an array.
def arrayMinimum(A,N):
# calculating minimum
# of first two numbers
mn = A[0]
# Iterating through each
# of the member of
# the array to calculate
# the minimum
for i in range(N-1,0,-1):
# Finding the minimum
# between current
# minimum and current value.
mn = minimum(mn, A[i])
return mn
# Driver code
A = [ 2, 3, 1, 4]
N =len(A)
print(arrayMinimum(A, N))
# This code is contributed
# by Anant Agarwal.
// C# program to find minimum in an
// array without using Relational Operators
using System;
class GFG
{
// Function to find minimum between two
// non-negative numbers without
// using relational operator.
static int minimum(int x, int y)
{
int c = 0;
// Continues till any
// element becomes zero
while (x > 0 && y > 0)
{
x--;
y--;
c++;
}
return c;
}
// Function to find minimum in an array.
static int arrayMinimum(int []A, int N)
{
// calculating minimum of
// first two numbers
int mn = A[0];
// Iterating through each of the
// member of the array to
// calculate the minimum
for (int i = N - 1; i > 0; i--)
// Finding the minimum between current
// minimum and current value.
mn = minimum(mn, A[i]);
return mn;
}
// Driver code
public static void Main()
{
int []A = {2, 3, 1, 4};
int N = A.Length;
Console.WriteLine(arrayMinimum(A, N));
}
}
// This code is contributed by vt_m.
<?php
// PHP program to find minimum
// in an array without using
// Relational Operators
// Function to find minimum
// between two non-negative
// numbers without using
// relational operator.
function minimum($x, $y)
{
$c = 0;
// Continues till any
// element becomes zero.
while ($x and $y)
{
$x--;
$y--;
$c++;
}
return $c;
}
// Function to find
// minimum in an array.
function arrayMinimum( $A, $N)
{
// calculating minimum of
// first two numbers
$mn = $A[0];
// Iterating through each
// of the member of the
// array to calculate
// the minimum
for ($i = $N - 1; $i; $i--)
// Finding the minimum
// between current minimum
// and current value.
$mn = minimum($mn, $A[$i]);
return $mn;
}
// Driver code
$A = array(2, 3, 1, 4);
$N = count($A);
echo arrayMinimum($A, $N);
// This code is contributed
// by anuj_67.
?>
<script>
// Javascript program to find minimum in an
// array without using Relational Operators
// Function to find minimum between two
// non-negative numbers without
// using relational operator.
function minimum(x, y)
{
let c = 0;
// Continues till any
// element becomes zero
while (x > 0 && y > 0)
{
x--;
y--;
c++;
}
return c;
}
// Function to find minimum in an array.
function arrayMinimum(A, N)
{
// calculating minimum of
// first two numbers
let mn = A[0];
// Iterating through each of the
// member of the array to
// calculate the minimum
for (let i = N - 1; i > 0; i--)
// Finding the minimum between current
// minimum and current value.
mn = minimum(mn, A[i]);
return mn;
}
let A = [2, 3, 1, 4];
let N = A.length;
document.write(arrayMinimum(A, N));
// This code is contributed by divyesh072019.
</script>
Output:
1
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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