Minimum index i such that all the elements from index i to given index are equal
Last Updated :
10 Apr, 2023
Given an array arr[] of integers and an integer pos, the task is to find the minimum index i such that all the elements from index i to index pos are equal.
Examples:
Input: arr[] = {2, 1, 1, 1, 5, 2}, pos = 3
Output: 1
Elements in index range [1, 3] are all equal to 1.
Input: arr[] = {2, 1, 1, 1, 5, 2}, pos = 5
Output: 5
Brute Force Approach:
A brute force approach to solve this problem is to use two nested loops. The outer loop iterates through all the indices from 0 to pos, and the inner loop checks if all the elements from the current index to pos are equal. If they are equal, the current index is the minimum required index, and the loops can be terminated.
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 minimum required index
int minIndex(int arr[], int n, int pos)
{
for (int i = 0; i <= pos; i++) {
int j;
for (j = i + 1; j <= pos; j++) {
if (arr[j] != arr[i]) {
break;
}
}
if (j == pos + 1) {
return i;
}
}
return -1; // If no such index exists
}
// Driver code
int main()
{
int arr[] = { 2, 1, 1, 1, 5, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
int pos = 4;
// Function Call
cout << minIndex(arr, n, pos);
return 0;
}
import java.util.*;
public class Main {
// Function to return the minimum required index
static int minIndex(int arr[], int n, int pos) {
for (int i = 0; i <= pos; i++) {
int j;
for (j = i + 1; j <= pos; j++) {
if (arr[j] != arr[i]) {
break;
}
}
if (j == pos + 1) {
return i;
}
}
return -1; // If no such index exists
}
// Driver code
public static void main(String[] args) {
int arr[] = { 2, 1, 1, 1, 5, 2 };
int n = arr.length;
int pos = 4;
// Function Call
System.out.println(minIndex(arr, n, pos));
}
}
// This code is contributed by Prajwal kandekar
// Function to return the minimum required index
function minIndex(arr, n, pos) {
for (let i = 0; i <= pos; i++) {
let j;
for (j = i + 1; j <= pos; j++) {
if (arr[j] !== arr[i]) {
break;
}
}
if (j === pos + 1) {
return i;
}
}
return -1; // If no such index exists
}
// Driver code
const arr = [2, 1, 1, 1, 5, 2];
const n = arr.length;
const pos = 4;
// Function call
console.log(minIndex(arr, n, pos));
def minIndex(arr, n, pos):
# Function to return the minimum required index
for i in range(pos+1):
j = i + 1
while j <= pos:
if arr[j] != arr[i]:
break
j += 1
if j == pos + 1:
return i
return -1 # If no such index exists
# Driver code
arr = [2, 1, 1, 1, 5, 2]
n = len(arr)
pos = 4
# Function call
print(minIndex(arr, n, pos))
using System;
class Program {
// Function to return the minimum required index
static int minIndex(int[] arr, int n, int pos) {
for (int i = 0; i <= pos; i++) {
int j;
for (j = i + 1; j <= pos; j++) {
if (arr[j] != arr[i]) {
break;
}
}
if (j == pos + 1) {
return i;
}
}
return -1; // If no such index exists
}
static void Main(string[] args) {
int[] arr = { 2, 1, 1, 1, 5, 2 };
int n = arr.Length;
int pos = 4;
// Function Call
Console.WriteLine(minIndex(arr, n, pos));
}
}
Output: 4
Time Complexity: O(n^2)
Space Complexity: O(1)
Simple Approach: Starting from index pos - 1, traverse the array in reverse and for the first index i such that arr[i] != arr[pos] print i + 1 which is the required index.
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 minimum required index
int minIndex(int arr[], int n, int pos)
{
int num = arr[pos];
// Start from arr[pos - 1]
int i = pos - 1;
while (i >= 0) {
if (arr[i] != num)
break;
i--;
}
// All elements are equal
// from arr[i + 1] to arr[pos]
return i + 1;
}
// Driver code
int main()
{
int arr[] = { 2, 1, 1, 1, 5, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
int pos = 4;
// Function Call
cout << minIndex(arr, n, pos);
return 0;
}
// Java implementation of the approach
import java.io.*;
public class GFG {
// Function to return the minimum required index
static int minIndex(int arr[], int n, int pos)
{
int num = arr[pos];
// Start from arr[pos - 1]
int i = pos - 1;
while (i >= 0) {
if (arr[i] != num)
break;
i--;
}
// All elements are equal
// from arr[i + 1] to arr[pos]
return i + 1;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 1, 1, 1, 5, 2 };
int n = arr.length;
int pos = 4;
// Function Call
System.out.println(minIndex(arr, n, pos));
}
}
// This code is contributed by Code_Mech.
# Python3 implementation of the approach
# Function to return the minimum
# required index
def minIndex(arr, n, pos):
num = arr[pos]
# Start from arr[pos - 1]
i = pos - 1
while (i >= 0):
if (arr[i] != num):
break
i -= 1
# All elements are equal
# from arr[i + 1] to arr[pos]
return i + 1
# Driver code
arr = [2, 1, 1, 1, 5, 2 ]
n = len(arr)
pos = 4
# Function Call
print(minIndex(arr, n, pos))
# This code is contributed by
# Mohit Kumar 29
// C# implementation of the approach
using System;
class GFG {
// Function to return the minimum required index
static int minIndex(int[] arr, int n, int pos)
{
int num = arr[pos];
// Start from arr[pos - 1]
int i = pos - 1;
while (i >= 0) {
if (arr[i] != num)
break;
i--;
}
// All elements are equal
// from arr[i + 1] to arr[pos]
return i + 1;
}
// Driver code
public static void Main()
{
int[] arr = { 2, 1, 1, 1, 5, 2 };
int n = arr.Length;
int pos = 4;
// Function Call
Console.WriteLine(minIndex(arr, n, pos));
}
}
// This code is contributed
// by Akanksha Rai
<?php
// PHP implementation of the approach
// Function to return the minimum
// required index
function minIndex($arr, $n, $pos)
{
$num = $arr[$pos];
// Start from arr[pos - 1]
$i = $pos - 1;
while ($i >= 0)
{
if ($arr[$i] != $num)
break;
$i--;
}
// All elements are equal
// from arr[i + 1] to arr[pos]
return $i + 1;
}
// Driver code
$arr = array(2, 1, 1, 1, 5, 2 );
$n = sizeof($arr);
$pos = 4;
echo minIndex($arr, $n, $pos);
// This code is contributed by Ryuga
?>
<script>
// Javascript implementation of the approach
// Function to return the minimum required index
function minIndex(arr, n, pos)
{
var num = arr[pos];
// Start from arr[pos - 1]
var i = pos - 1;
while (i >= 0) {
if (arr[i] != num)
break;
i--;
}
// All elements are equal
// from arr[i + 1] to arr[pos]
return i + 1;
}
// Driver code
var arr = [ 2, 1, 1, 1, 5, 2 ];
var n = arr.length;
var pos = 4;
// Function Call
document.write(minIndex(arr, n, pos));
// This code is contributed by rrrtnx.
</script>
Time Complexity: O(N)
Space Complexity: O(1)
Efficient Approach :
Do a binary search in the sub-array [0, pos-1]. Stop condition will be if arr[mid] == arr[pos] && arr[mid-1] != arr[pos]. Go-left or Go-right will depend on if arr[mid] == arr[pos] or not respectively.
Implementation:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the minimum required index
int minIndex(int arr[], int pos)
{
int low = 0;
int high = pos;
int i = pos;
while (low < high) {
int mid = (low + high) / 2;
if (arr[mid] != arr[pos]) {
low = mid + 1;
}
else {
high = mid - 1;
i = mid;
if (mid > 0 && arr[mid - 1] != arr[pos]) {
// Short-circuit more comparisons as found
// the border point
break;
}
}
}
// For cases were high = low + 1 and arr[high] will
// match with
// arr[pos] but not arr[low] or arr[mid]. In such
// iteration the if condition will satisfy and loop will
// break post that low will be updated. Hence i will not
// point to the correct index.
return arr[low] == arr[pos] ? low : i;
}
// Driver code
int main()
{
int arr[] = { 2, 1, 1, 1, 5, 2 };
cout << minIndex(arr, 2) << endl; // Should be 1
cout << minIndex(arr, 3) << endl; // Should be 1
cout << minIndex(arr, 4) << endl; // Should be 4
return 0;
}
// This code is contributed by
// anshbikram
// Java implementation of the approach
public class GFG {
// Function to return the minimum required index
static int minIndex(int arr[], int pos)
{
int low = 0;
int high = pos;
int i = pos;
while (low < high) {
int mid = (low + high) / 2;
if (arr[mid] != arr[pos]) {
low = mid + 1;
}
else {
high = mid - 1;
i = mid;
if (mid > 0 && arr[mid - 1] != arr[pos]) {
// Short-circuit more comparisons as
// found the border point
break;
}
}
}
// For cases were high = low + 1 and arr[high] will
// match with arr[pos] but not arr[low] or arr[mid].
// In such iteration the if condition will satisfy
// and loop will break post that low will be
// updated. Hence i will not point to the correct
// index.
return arr[low] == arr[pos] ? low : i;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 1, 1, 1, 5, 2 };
System.out.println(minIndex(arr, 2)); // Should be 1
System.out.println(minIndex(arr, 3)); // Should be 1
System.out.println(minIndex(arr, 4)); // Should be 4
}
}
// This code is contributed by
// anshbikram
# Python3 implementation of the approach
# Function to return the minimum
# required index
def minIndex(arr, pos):
low = 0
high = pos
i = pos
while low < high:
mid = (low + high)//2
if arr[mid] != arr[pos]:
low = mid + 1
else:
high = mid - 1
i = mid
if mid > 0 and arr[mid-1] != arr[pos]:
# Short-circuit more comparisons as found the border point
break
# For cases were high = low + 1 and arr[high] will match with
# arr[pos] but not arr[low] or arr[mid]. In such iteration
# the if condition will satisfy and loop will break post that
# low will be updated. Hence i will not point to the correct index.
return low if arr[low] == arr[pos] else i
# Driver code
arr = [2, 1, 1, 1, 5, 2]
print(minIndex(arr, 2)) # Should be 1
print(minIndex(arr, 3)) # Should be 1
print(minIndex(arr, 4)) # Should be 4
# This code is contributed by
# anshbikram
// C# implementation of the approach
using System;
class GFG{
// Function to return the minimum
// required index
static int minIndex(int []arr, int pos)
{
int low = 0;
int high = pos;
int i = pos;
while (low < high)
{
int mid = (low + high) / 2;
if (arr[mid] != arr[pos])
{
low = mid + 1;
}
else
{
high = mid - 1;
i = mid;
if (mid > 0 && arr[mid - 1] != arr[pos])
{
// Short-circuit more comparisons as
// found the border point
break;
}
}
}
// For cases were high = low + 1 and arr[high] will
// match with arr[pos] but not arr[low] or arr[mid].
// In such iteration the if condition will satisfy
// and loop will break post that low will be
// updated. Hence i will not point to the correct
// index.
return arr[low] == arr[pos] ? low : i;
}
// Driver code
public static void Main()
{
int []arr = { 2, 1, 1, 1, 5, 2 };
Console.WriteLine(minIndex(arr, 2)); // Should be 1
Console.WriteLine(minIndex(arr, 3)); // Should be 1
Console.WriteLine(minIndex(arr, 4)); // Should be 4
}
}
// This code is contributed by chitranayal
<script>
// javascript implementation of the approach
// Function to return the minimum required index
function minIndex(arr , pos) {
var low = 0;
var high = pos;
var i = pos;
while (low < high) {
var mid = parseInt((low + high) / 2);
if (arr[mid] != arr[pos]) {
low = mid + 1;
} else {
high = mid - 1;
i = mid;
if (mid > 0 && arr[mid - 1] != arr[pos]) {
// Short-circuit more comparisons as
// found the border point
break;
}
}
}
// For cases were high = low + 1 and arr[high] will
// match with arr[pos] but not arr[low] or arr[mid].
// In such iteration the if condition will satisfy
// and loop will break post that low will be
// updated. Hence i will not point to the correct
// index.
return arr[low] == arr[pos] ? low : i;
}
// Driver code
var arr = [ 2, 1, 1, 1, 5, 2 ];
document.write(minIndex(arr, 2)+"<br/>"); // Should be 1
document.write(minIndex(arr, 3)+"<br/>"); // Should be 1
document.write(minIndex(arr, 4)+"<br/>"); // Should be 4
// This code is contributed by todaysgaurav
</script>
Time Complexity: O(log(n))
Space Complexity: O(1)
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