Leftmost and rightmost indices of the maximum and the minimum element of an array
Last Updated :
11 Jul, 2025
Given an array arr[], the task is to find the leftmost and the rightmost indices of the minimum and the maximum element from the array where arr[] consists of non-distinct elements.
Examples:
Input: arr[] = {2, 1, 1, 2, 1, 5, 6, 5}
Output: Minimum left : 1
Minimum right : 4
Maximum left : 6
Maximum right : 6
Minimum element is 1 which is present at indices 1, 2 and 4.
Maximum element is 6 which is present only at index 6.
Input: arr[] = {0, 1, 0, 2, 7, 5, 6, 7}
Output: Minimum left : 0
Minimum right : 2
Maximum left : 4
Maximum right : 7
Method 1: When the array is unsorted.
- Initialize the variable leftMin = rightMin = leftMax = rightMax = arr[0] and min = max = arr[0].
- Start traversing the array from 1 to n - 1.
- If arr[i] < min then a new minimum is found. Update leftMin = rightMin = i.
- Else arr[i] = min then another copy of the current minimum is found. Update the rightMin = i.
- If arr[i] > max then a new maximum is found. Update leftMax = rightMax = i.
- Else arr[i] = max then another copy of the current maximum is found. Update the rightMax = i.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
void findIndices(int arr[], int n)
{
int leftMin = 0, rightMin = 0;
int leftMax = 0, rightMax = 0;
int min = arr[0], max = arr[0];
for (int i = 1; i < n; i++) {
// If found new minimum
if (arr[i] < min) {
leftMin = rightMin = i;
min = arr[i];
}
// If arr[i] = min then rightmost index
// for min will change
else if (arr[i] == min)
rightMin = i;
// If found new maximum
if (arr[i] > max) {
leftMax = rightMax = i;
max = arr[i];
}
// If arr[i] = max then rightmost index
// for max will change
else if (arr[i] == max)
rightMax = i;
}
cout << "Minimum left : " << leftMin << "\n";
cout << "Minimum right : " << rightMin <<"\n";
cout << "Maximum left : " << leftMax <<"\n";
cout << "Maximum right : " << rightMax <<"\n";
}
// Driver code
int main()
{
int arr[] = { 2, 1, 1, 2, 1, 5, 6, 5 };
int n = sizeof(arr)/sizeof(arr[0]);
findIndices(arr, n);
}
// This code is contributed
// by ihritik
// Java implementation of the approach
public class GFG {
public static void findIndices(int arr[], int n)
{
int leftMin = 0, rightMin = 0;
int leftMax = 0, rightMax = 0;
int min = arr[0], max = arr[0];
for (int i = 1; i < n; i++) {
// If found new minimum
if (arr[i] < min) {
leftMin = rightMin = i;
min = arr[i];
}
// If arr[i] = min then rightmost index
// for min will change
else if (arr[i] == min)
rightMin = i;
// If found new maximum
if (arr[i] > max) {
leftMax = rightMax = i;
max = arr[i];
}
// If arr[i] = max then rightmost index
// for max will change
else if (arr[i] == max)
rightMax = i;
}
System.out.println("Minimum left : " + leftMin);
System.out.println("Minimum right : " + rightMin);
System.out.println("Maximum left : " + leftMax);
System.out.println("Maximum right : " + rightMax);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 1, 1, 2, 1, 5, 6, 5 };
int n = arr.length;
findIndices(arr, n);
}
}
# Python3 implementation of the approach
def findIndices(arr, n) :
leftMin, rightMin = 0, 0
leftMax, rightMax = 0, 0
min_element = arr[0]
max_element = arr[0]
for i in range(n) :
# If found new minimum
if (arr[i] < min_element) :
leftMin = rightMin = i
min_element = arr[i]
# If arr[i] = min then rightmost
# index for min will change
elif (arr[i] == min_element) :
rightMin = i
# If found new maximum
if (arr[i] > max_element) :
leftMax = rightMax = i
max_element = arr[i]
# If arr[i] = max then rightmost
# index for max will change
elif (arr[i] == max_element) :
rightMax = i
print("Minimum left : ", leftMin)
print("Minimum right : ", rightMin)
print("Maximum left : ", leftMax )
print("Maximum right : ", rightMax)
# Driver code
if __name__ == "__main__" :
arr = [ 2, 1, 1, 2, 1, 5, 6, 5 ]
n = len(arr)
findIndices(arr, n)
# This code is contributed by Ryuga
// C# implementation of the approach
using System;
class GFG {
static void findIndices(int []arr, int n)
{
int leftMin = 0, rightMin = 0;
int leftMax = 0, rightMax = 0;
int min = arr[0], max = arr[0];
for (int i = 1; i < n; i++) {
// If found new minimum
if (arr[i] < min) {
leftMin = rightMin = i;
min = arr[i];
}
// If arr[i] = min then rightmost index
// for min will change
else if (arr[i] == min)
rightMin = i;
// If found new maximum
if (arr[i] > max) {
leftMax = rightMax = i;
max = arr[i];
}
// If arr[i] = max then rightmost index
// for max will change
else if (arr[i] == max)
rightMax = i;
}
Console.WriteLine("Minimum left : " + leftMin);
Console.WriteLine("Minimum right : " + rightMin);
Console.WriteLine("Maximum left : " + leftMax);
Console.WriteLine("Maximum right : " + rightMax);
}
// Driver code
public static void Main()
{
int []arr = { 2, 1, 1, 2, 1, 5, 6, 5 };
int n = arr.Length;
findIndices(arr, n);
}
}
// This code is contributed
// By ihritik
<?php
// PHP implementation of the approach
function findIndices($arr, $n)
{
$leftMin = 0;
$rightMin = 0;
$leftMax = 0;
$rightMax = 0;
$min = $arr[0];
$max = $arr[0];
for ($i = 1; $i < $n; $i++)
{
// If found new minimum
if ($arr[$i] < $min)
{
$leftMin = $rightMin = $i;
$min = $arr[$i];
}
// If arr[i] = min then rightmost
// index for min will change
else if ($arr[$i] == $min)
$rightMin = $i;
// If found new maximum
if ($arr[$i] > $max)
{
$leftMax = $rightMax = $i;
$max = $arr[$i];
}
// If arr[i] = max then rightmost
// index for max will change
else if ($arr[$i] == $max)
$rightMax = $i;
}
echo "Minimum left : ", $leftMin, "\n";
echo "Minimum right : ", $rightMin,"\n";
echo "Maximum left : ", $leftMax, "\n";
echo "Maximum right : ", $rightMax, "\n";
}
// Driver code
$arr = array( 2, 1, 1, 2, 1, 5, 6, 5 );
$n = sizeof($arr);
findIndices($arr, $n);
// This code is contributed
// by Sachin
?>
<script>
// Javascript implementation of the approach
function findIndices(arr,n)
{
let leftMin = 0, rightMin = 0;
let leftMax = 0, rightMax = 0;
let min = arr[0], max = arr[0];
for (let i = 1; i < n; i++) {
// If found new minimum
if (arr[i] < min) {
leftMin = rightMin = i;
min = arr[i];
}
// If arr[i] = min then rightmost index
// for min will change
else if (arr[i] == min)
rightMin = i;
// If found new maximum
if (arr[i] > max) {
leftMax = rightMax = i;
max = arr[i];
}
// If arr[i] = max then rightmost index
// for max will change
else if (arr[i] == max)
rightMax = i;
}
document.write("Minimum left : " + leftMin+"<br>");
document.write("Minimum right : " + rightMin+"<br>");
document.write("Maximum left : " + leftMax+"<br>");
document.write("Maximum right : " + rightMax+"<br>");
}
// Driver code
let arr=[2, 1, 1, 2, 1, 5, 6, 5 ];
let n = arr.length;
findIndices(arr, n);
// This code is contributed by unknown2108
</script>
Output: Minimum left : 1
Minimum right : 4
Maximum left : 6
Maximum right : 6
Time Complexity: O(n)
Auxiliary Space: O(1)
Method 2: When the array is sorted.
- When the array is sorted then leftMin = 0 and rightMax = n - 1.
- In order to find the rightMin, apply a modified binary search:
- Set i = 1.
- While arr[i] = min update rightMin = i and i = i * 2.
- Finally do a linear search for the rest of the elements from rightMin + 1 to n - 1 while arr[i] = min.
- Return rightMin in the end.
- Similarly, for leftMax repeat the above steps but in reverse i.e. from n - 1 and update i = i / 2 after every iteration.
Below is the implementation of the above approach:
C++
// C++ implementation of above idea
#include<bits/stdc++.h>
using namespace std;
// Function to return the index of the rightmost
// minimum element from the array
int getRightMin(int arr[], int n)
{
// First element is the minimum in a sorted array
int min = arr[0];
int rightMin = 0;
int i = 1;
while (i < n) {
// While the elements are equal to the minimum
// update rightMin
if (arr[i] == min)
rightMin = i;
i *= 2;
}
i = rightMin + 1;
// Final check whether there are any elements
// which are equal to the minimum
while (i < n && arr[i] == min) {
rightMin = i;
i++;
}
return rightMin;
}
// Function to return the index of the leftmost
// maximum element from the array
int getLeftMax(int arr[], int n)
{
// Last element is the maximum in a sorted array
int max = arr[n - 1];
int leftMax = n - 1;
int i = n - 2;
while (i > 0) {
// While the elements are equal to the maximum
// update leftMax
if (arr[i] == max)
leftMax = i;
i /= 2;
}
i = leftMax - 1;
// Final check whether there are any elements
// which are equal to the maximum
while (i >= 0 && arr[i] == max) {
leftMax = i;
i--;
}
return leftMax;
}
// Driver code
int main()
{
int arr[] = { 0, 0, 1, 2, 5, 5, 6, 8, 8 };
int n = sizeof(arr)/sizeof(arr[0]);
// First element is the leftmost minimum in a sorted array
cout << "Minimum left : " << 0 <<"\n";
cout << "Minimum right : " << getRightMin(arr, n) << "\n";
cout << "Maximum left : " << getLeftMax(arr, n) <<"\n";
// Last element is the rightmost maximum in a sorted array
cout << "Maximum right : " << (n - 1);
}
// This code is contributed by ihritik
// Java implementation of above idea
public class GFG {
// Function to return the index of the rightmost
// minimum element from the array
public static int getRightMin(int arr[], int n)
{
// First element is the minimum in a sorted array
int min = arr[0];
int rightMin = 0;
int i = 1;
while (i < n) {
// While the elements are equal to the minimum
// update rightMin
if (arr[i] == min)
rightMin = i;
i *= 2;
}
i = rightMin + 1;
// Final check whether there are any elements
// which are equal to the minimum
while (i < n && arr[i] == min) {
rightMin = i;
i++;
}
return rightMin;
}
// Function to return the index of the leftmost
// maximum element from the array
public static int getLeftMax(int arr[], int n)
{
// Last element is the maximum in a sorted array
int max = arr[n - 1];
int leftMax = n - 1;
int i = n - 2;
while (i > 0) {
// While the elements are equal to the maximum
// update leftMax
if (arr[i] == max)
leftMax = i;
i /= 2;
}
i = leftMax - 1;
// Final check whether there are any elements
// which are equal to the maximum
while (i >= 0 && arr[i] == max) {
leftMax = i;
i--;
}
return leftMax;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 0, 0, 1, 2, 5, 5, 6, 8, 8 };
int n = arr.length;
// First element is the leftmost minimum in a sorted array
System.out.println("Minimum left : " + 0);
System.out.println("Minimum right : " + getRightMin(arr, n));
System.out.println("Maximum left : " + getLeftMax(arr, n));
// Last element is the rightmost maximum in a sorted array
System.out.println("Maximum right : " + (n - 1));
}
}
# Python 3 implementation of above idea
# Function to return the index of the
# rightmost minimum element from the array
def getRightMin(arr, n):
# First element is the minimum
# in a sorted array
min = arr[0]
rightMin = 0
i = 1
while (i < n):
# While the elements are equal to
# the minimum update rightMin
if (arr[i] == min):
rightMin = i
i *= 2
i = rightMin + 1
# Final check whether there are any
# elements which are equal to the minimum
while (i < n and arr[i] == min):
rightMin = i
i += 1
return rightMin
# Function to return the index of the
# leftmost maximum element from the array
def getLeftMax(arr, n):
# Last element is the maximum
# in a sorted array
max = arr[n - 1]
leftMax = n - 1
i = n - 2
while (i > 0):
# While the elements are equal to
# the maximum update leftMax
if (arr[i] == max):
leftMax = i
i = int(i / 2)
i = leftMax - 1
# Final check whether there are any
# elements which are equal to the maximum
while (i >= 0 and arr[i] == max):
leftMax = i
i -= 1
return leftMax
# Driver code
if __name__ == '__main__':
arr = [0, 0, 1, 2, 5, 5, 6, 8, 8]
n = len(arr)
# First element is the leftmost
# minimum in a sorted array
print("Minimum left :", 0)
print("Minimum right :", getRightMin(arr, n))
print("Maximum left :", getLeftMax(arr, n))
# Last element is the rightmost maximum
# in a sorted array
print("Maximum right :", (n - 1))
# This code is contributed by
# Surendra_Gangwar
// C# implementation of above idea
using System;
public class GFG {
// Function to return the index of the rightmost
// minimum element from the array
public static int getRightMin(int []arr, int n)
{
// First element is the minimum in a sorted array
int min = arr[0];
int rightMin = 0;
int i = 1;
while (i < n) {
// While the elements are equal to the minimum
// update rightMin
if (arr[i] == min)
rightMin = i;
i *= 2;
}
i = rightMin + 1;
// Final check whether there are any elements
// which are equal to the minimum
while (i < n && arr[i] == min) {
rightMin = i;
i++;
}
return rightMin;
}
// Function to return the index of the leftmost
// maximum element from the array
public static int getLeftMax(int []arr, int n)
{
// Last element is the maximum in a sorted array
int max = arr[n - 1];
int leftMax = n - 1;
int i = n - 2;
while (i > 0) {
// While the elements are equal to the maximum
// update leftMax
if (arr[i] == max)
leftMax = i;
i /= 2;
}
i = leftMax - 1;
// Final check whether there are any elements
// which are equal to the maximum
while (i >= 0 && arr[i] == max) {
leftMax = i;
i--;
}
return leftMax;
}
// Driver code
public static void Main()
{
int []arr = { 0, 0, 1, 2, 5, 5, 6, 8, 8 };
int n = arr.Length;
// First element is the leftmost minimum in a sorted array
Console.WriteLine("Minimum left : " + 0);
Console.WriteLine("Minimum right : " + getRightMin(arr, n));
Console.WriteLine("Maximum left : " + getLeftMax(arr, n));
// Last element is the rightmost maximum in a sorted array
Console.WriteLine("Maximum right : " + (n - 1));
}
}
// This code is contributed by ihritik
<?php
// PHP implementation of above idea
// Function to return the index of the
// rightmost minimum element from the array
function getRightMin($arr, $n)
{
// First element is the minimum
// in a sorted array
$min = $arr[0];
$rightMin = 0;
$i = 1;
while ($i < $n)
{
// While the elements are equal to
// the minimum update rightMin
if ($arr[$i] == $min)
$rightMin = $i;
$i *= 2;
}
$i = $rightMin + 1;
// Final check whether there are any
// elements which are equal to the minimum
while ($i < $n && $arr[$i] == $min)
{
$rightMin = $i;
$i++;
}
return $rightMin;
}
// Function to return the index of the
// leftmost maximum element from the array
function getLeftMax($arr, $n)
{
// Last element is the maximum in
// a sorted array
$max = $arr[$n - 1];
$leftMax = $n - 1;
$i = $n - 2;
while ($i > 0)
{
// While the elements are equal to
// the maximum update leftMax
if ($arr[$i] == $max)
$leftMax = $i;
$i /= 2;
}
$i = $leftMax - 1;
// Final check whether there are any
// elements which are equal to the maximum
while ($i >= 0 && $arr[$i] == $max)
{
$leftMax = $i;
$i--;
}
return $leftMax;
}
// Driver code
$arr = array(0, 0, 1, 2, 5,
5, 6, 8, 8 );
$n = sizeof($arr);
// First element is the leftmost
// minimum in a sorted array
echo "Minimum left : ", 0, "\n";
echo "Minimum right : ",
getRightMin($arr, $n), "\n";
echo "Maximum left : ",
getLeftMax($arr, $n), "\n";
// Last element is the rightmost
// maximum in a sorted array
echo "Maximum right : ", ($n - 1), "\n";
// This code is Contributed
// by Mukul singh
?>
<script>
// Javascript implementation of above idea
// Function to return the index of the rightmost
// minimum element from the array
function getRightMin(arr, n)
{
// First element is the minimum in a sorted array
let min = arr[0];
let rightMin = 0;
let i = 1;
while (i < n) {
// While the elements are equal to the minimum
// update rightMin
if (arr[i] == min)
rightMin = i;
i *= 2;
}
i = rightMin + 1;
// Final check whether there are any elements
// which are equal to the minimum
while (i < n && arr[i] == min) {
rightMin = i;
i++;
}
return rightMin;
}
// Function to return the index of the leftmost
// maximum element from the array
function getLeftMax(arr, n)
{
// Last element is the maximum in a sorted array
let max = arr[n - 1];
let leftMax = n - 1;
let i = n - 2;
while (i > 0) {
// While the elements are equal to the maximum
// update leftMax
if (arr[i] == max)
leftMax = i;
i = parseInt(i / 2, 10);
}
i = leftMax - 1;
// Final check whether there are any elements
// which are equal to the maximum
while (i >= 0 && arr[i] == max) {
leftMax = i;
i--;
}
return leftMax;
}
let arr = [ 0, 0, 1, 2, 5, 5, 6, 8, 8 ];
let n = arr.length;
// First element is the leftmost minimum in a sorted array
document.write("Minimum left : " + 0 + "</br>");
document.write("Minimum right : " + getRightMin(arr, n) + "</br>");
document.write("Maximum left : " + getLeftMax(arr, n) + "</br>");
// Last element is the rightmost maximum in a sorted array
document.write("Maximum right : " + (n - 1));
// This code is contributed by suresh07.
</script>
Output: Minimum left : 0
Minimum right : 1
Maximum left : 7
Maximum right : 8
Time Complexity: O(n) As linear search is applied for a set of elements so the worst case time complexity will be O(n).
Auxiliary Space: O(1)
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