Given an array of integers, task is to find the starting and ending position of a given key.
Examples:
Input : arr[] = {1, 2, 3, 4, 5, 5}
Key = 5
Output : Start index: 4
Last index: 5
Explanation: Starting index where 5
is present is 4 and ending index is 5.
Input :arr[] = {1, 3, 7, 8, 6},
Key = 2
Output : Key not present in array
Input :arr[] = {1, 8, 7, 8, 6},
Key = 7
Output : Only one occurrence of
key is present at index 2We traverse array from beginning to find first occurrence. If element is present, then we traverse from end also to find last occurrence.
Implementation:
C++
// CPP program to find starting and ending // indexes of repeated numbers in an array #include <iostream> using namespace std;
// Function to find starting and end index void findIndex(int a[], int n, int key)
{ int start = -1;
// Traverse from beginning to find
// first occurrence
for (int i = 0; i < n; i++) {
if (a[i] == key) {
start = i;
break;
}
}
if (start == -1) {
cout << "Key not present in array";
return;
}
// Traverse from end to find last
// occurrence.
int end = start;
for (int i = n - 1; i >= start; i--) {
if (a[i] == key) {
end = i;
break;
}
}
if (start == end)
cout << "Only one key is present at index : "
<< start;
else {
cout << "Start index: " << start;
cout << "\n";
cout << "Last index: " << end;
}
} // Driver Code int main()
{ int a[] = { 1, 2, 7, 8, 8, 9, 8, 0, 0, 0, 8 };
int n = sizeof(a) / sizeof(a[0]);
// Key to find
int key = 8;
// Calling function
findIndex(a, n, key);
return 0;
} |
Java
// Java program to find starting and ending // indexes of repeated numbers in an array class Test {
// Function to find starting and end index
static void findIndex(int a[], int n, int key)
{
int start = -1;
// Traverse from beginning to find
// first occurrence
for (int i = 0; i < n; i++) {
if (a[i] == key) {
start = i;
break;
}
}
if (start == -1) {
System.out.println("Key not present in array");
return;
}
// Traverse from end to find last
// occurrence.
int end = start;
for (int i = n - 1; i >= start; i--) {
if (a[i] == key) {
end = i;
break;
}
}
if (start == end)
System.out.println("Only one key is present at index : " + start);
else {
System.out.println("Start index: " + start);
System.out.println("Last index: " + end);
}
}
// Driver method
public static void main(String args[])
{
int a[] = { 1, 2, 7, 8, 8, 9, 8, 0, 0, 0, 8 };
// Key to find
int key = 8;
// Calling method
findIndex(a, a.length, key);
}
} |
Python3
# Python3 code to find starting and ending # indexes of repeated numbers in an array # Function to find starting and end index def findIndex (a, n, key ):
start = -1
# Traverse from beginning to find
# first occurrence
for i in range(n):
if a[i] == key:
start = i
break
if start == -1:
print("Key not present in array")
return 0
# Traverse from end to find last
# occurrence.
end = start
for i in range(n-1, start - 1, -1):
if a[i] == key:
end = i
break
if start == end:
print("Only one key is present at index : ", start)
else:
print("Start index: ", start)
print("Last index: ", end)
# Driver Code a = [1, 2, 7, 8, 8, 9, 8, 0, 0, 0, 8]
n = len(a)
# Key to find key = 8
# Calling function findIndex(a, n, key) # This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program to find starting and ending // indexes of repeated numbers in an array using System;
class GFG {
// Function to find starting and
// end index
static void findIndex(int[] a, int n,
int key)
{
int start = -1;
// Traverse from beginning to
// find first occurrence
for (int i = 0; i < n; i++) {
if (a[i] == key) {
start = i;
break;
}
}
if (start == -1) {
Console.WriteLine("Key not "
+ "present in array");
return;
}
// Traverse from end to find last
// occurrence.
int end = start;
for (int i = n - 1; i >= start; i--) {
if (a[i] == key) {
end = i;
break;
}
}
if (start == end)
Console.WriteLine("Only one key is"
+ " present at index : "
+ start);
else {
Console.WriteLine("Start index: "
+ start);
Console.WriteLine("Last index: "
+ end);
}
}
// Driver method
public static void Main()
{
int[] a = { 1, 2, 7, 8, 8, 9,
8, 0, 0, 0, 8 };
// Key to find
int key = 8;
// Calling method
findIndex(a, a.Length, key);
}
} // This code is contributed by parashar. |
PHP
<?php // PHP program to find starting and ending // indexes of repeated numbers in an array // Function to find starting and end index function findIndex($arr, $find){
// To store the atrting and
// the ending index of the key
$start = -1;
$end = -1;
// For every element of the given array
foreach ($arr as $key => $value) {
// If current element is equal
// to the given key
if($find === $value){
// If starting index hasn't been set
if($start==-1)
$start = $key;
$end = $key;
}
}
// If key is not present in the array
if($start==-1){
return "Key not present in array";
}
if($end == $start){
return "Only one key is present "." at index : ". $start;
}
return "Start index: ".$start. "\nLast index: ".$end;
} // Driver code $a = array(1, 2, 7, 8, 8, 9, 8, 0, 0, 0, 8);
// Key to find $key = 8;
// Calling function echo findIndex($a, $key);
?> |
Javascript
<script> // Javascript program to find starting and ending // indexes of repeated numbers in an array // Function to find starting and end index function findIndex(a, n, key)
{ let start = -1;
// Traverse from beginning to find
// first occurrence
for (let i = 0; i < n; i++) {
if (a[i] == key) {
start = i;
break;
}
}
if (start == -1) {
document.write("Key not present in array");
return;
}
// Traverse from end to find last
// occurrence.
let end = start;
for (let i = n - 1; i >= start; i--) {
if (a[i] == key) {
end = i;
break;
}
}
if (start == end)
document.write("Only one key is present at index : "
+ start);
else {
document.write("Start index: " + start);
document.write("<br>" + "Last index: " + end);
}
} // Driver Code let a = [ 1, 2, 7, 8, 8, 9, 8, 0, 0, 0, 8 ];
let n = a.length;
// Key to find
let key = 8;
// Calling function
findIndex(a, n, key);
//This code is contributed by Mayank Tyagi </script> |
Output
Start index: 3 Last index: 10
Time Complexity: Worst case time complexity is O(N), ( when we traverse the whole array and don’t find the element’s start and last indices), where N represents the size of the given array. and best case time complexity will be O(1), when start index is ‘0’ and last index is ‘n – 1’.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Related Article:
Find first and last occurrences of an element in a sorted array