Given an array arr[] of size N. For every element in the array, the task is to find the index of the farthest element in the array to the right which is smaller than the current element. If no such number exists then print -1.
Examples:
Input: arr[] = {3, 1, 5, 2, 4}
Output: 3 -1 4 -1 -1
arr[3] is the farthest smallest element to the right of arr[0].
arr[4] is the farthest smallest element to the right of arr[2].
And for the rest of the elements, there is no smaller element to their right.Input: arr[] = {1, 2, 3, 4, 0}
Output: 4 4 4 4 -1
Approach 1 : (Brute Force Method)
A brute force approach to this problem can be, keep a variable idx = -1 from beginning and for each element start traversing the same array from the backward upto (i+1)th index. And, if at any index j find smaller element from the current element, i.e. (a[i] > a[j]) break from the loop.
Below is the implementation of the above approach :
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to find the farthest // smaller number in the right side void farthest_min( int a[], int n)
{ for ( int i = 0; i < n; i++) {
// keeping the idx = -1 from beginning
int idx = -1;
// traverse the given array backward
for ( int j = n - 1; j > i; j--) {
// if found any element smaller
if (a[i] > a[j]) {
// update that index and break
idx = j;
break ;
}
}
// Print the required index
cout << idx << " " ;
}
} // Driver code int main()
{ int a[] = { 3, 1, 5, 2, 4 };
int n = sizeof (a) / sizeof (a[0]);
farthest_min(a, n);
return 0;
} // this code is contributed by Rajdeep |
// Java implementation of the approach public class GFG {
// Function to find the farthest
// smaller number in the right side
static void farthest_min( int [] a, int n)
{
// To store minimum element
// in the range i to n
for ( int i = 0 ; i < n; i++) {
// keeping the idx = -1 from beginning
int idx = - 1 ;
for ( int j = n - 1 ; j > i; j--) {
// if found any element smaller
if (a[i] > a[j]) {
// update that index and break
idx = j;
break ;
}
}
System.out.print(idx + " " );
}
}
// Driver code
public static void main(String[] args)
{
int [] a = { 3 , 1 , 5 , 2 , 4 };
int n = a.length;
farthest_min(a, n);
}
} // This code is contributed by Rajdeep |
# Python 3 implementation of the approach # Function to find the farthest # smaller number in the right side def farthest_min(a, n):
for i in range (n):
# keeping the idx = -1 from beginning
idx = - 1
j = n - 1
# traverse the given array backward
while j > i:
# if found any element smaller
if a[i] > a[j]:
# update that index and break
idx = j
break
j - = 1
# Print the required index
print (idx,end = " " )
# Driver code a = [ 3 , 1 , 5 , 2 , 4 ]
n = len (a)
farthest_min(a, n) "This Code is contributed by rajatkumargla19" |
// C# implementation of the approach using System;
class GFG {
// Function to find the farthest
// smaller number in the right side
static void farthest_min( int [] a, int n)
{
// To store minimum element
// in the range i to n
for ( int i = 0; i < n; i++) {
// keeping the idx = -1 from beginning
int idx = -1;
for ( int j = n - 1; j > i; j--) {
// if found any element smaller
if (a[i] > a[j]) {
// update that index and break
idx = j;
break ;
}
}
Console.Write(idx + " " );
}
}
// Driver code
public static void Main()
{
int [] a = { 3, 1, 5, 2, 4 };
int n = a.Length;
farthest_min(a, n);
}
} // This code is contributed by Samim Hossain Mondal. |
function farthest_min(a, n)
{
// To store minimum element
// in the range i to n
for (let i = 0; i < n; i++) {
// keeping the idx = -1 from beginning
let idx = -1;
for (let j = n - 1; j > i; j--) {
// if found any element smaller
if (a[i] > a[j]) {
// update that index and break
idx = j;
break ;
}
}
console.log(idx + " " );
}
}
let a = [ 3, 1, 5, 2, 4 ];
let n = a.length;
farthest_min(a, n);
// This code is contributed by aadityaburujwale.
|
3 -1 4 -1 -1
Time Complexity : O(N^2)
Auxiliary Space : O(1)
Approach 2 : (Optimized Method)
An efficient approach is to create a suffix_min[] array where suffix_min[i] stores the minimum element from the subarray arr[i … N – 1]. Now for any element arr[i], binary search can be used on the subarray suffix_min[i + 1 … N – 1] to find the farthest smallest element to the right of arr[i].
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to find the farthest // smaller number in the right side void farthest_min( int a[], int n)
{ // To store minimum element
// in the range i to n
int suffix_min[n];
suffix_min[n - 1] = a[n - 1];
for ( int i = n - 2; i >= 0; i--) {
suffix_min[i] = min(suffix_min[i + 1], a[i]);
}
for ( int i = 0; i < n; i++) {
int low = i + 1, high = n - 1, ans = -1;
while (low <= high) {
int mid = (low + high) / 2;
// If current element in the suffix_min
// is less than a[i] then move right
if (suffix_min[mid] < a[i]) {
ans = mid;
low = mid + 1;
}
else
high = mid - 1;
}
// Print the required answer
cout << ans << " " ;
}
} // Driver code int main()
{ int a[] = { 3, 1, 5, 2, 4 };
int n = sizeof (a) / sizeof (a[0]);
farthest_min(a, n);
return 0;
} |
// Java implementation of the approach class GFG {
// Function to find the farthest
// smaller number in the right side
static void farthest_min( int [] a, int n)
{
// To store minimum element
// in the range i to n
int [] suffix_min = new int [n];
suffix_min[n - 1 ] = a[n - 1 ];
for ( int i = n - 2 ; i >= 0 ; i--) {
suffix_min[i]
= Math.min(suffix_min[i + 1 ], a[i]);
}
for ( int i = 0 ; i < n; i++) {
int low = i + 1 , high = n - 1 , ans = - 1 ;
while (low <= high) {
int mid = (low + high) / 2 ;
// If current element in the suffix_min
// is less than a[i] then move right
if (suffix_min[mid] < a[i]) {
ans = mid;
low = mid + 1 ;
}
else
high = mid - 1 ;
}
// Print the required answer
System.out.print(ans + " " );
}
}
// Driver code
public static void main(String[] args)
{
int [] a = { 3 , 1 , 5 , 2 , 4 };
int n = a.length;
farthest_min(a, n);
}
} // This code is contributed by ihritik |
# Python3 implementation of the approach # Function to find the farthest # smaller number in the right side def farthest_min(a, n):
# To store minimum element
# in the range i to n
suffix_min = [ 0 for i in range (n)]
suffix_min[n - 1 ] = a[n - 1 ]
for i in range (n - 2 , - 1 , - 1 ):
suffix_min[i] = min (suffix_min[i + 1 ], a[i])
for i in range (n):
low = i + 1
high = n - 1
ans = - 1
while (low < = high):
mid = (low + high) / / 2
# If current element in the suffix_min
# is less than a[i] then move right
if (suffix_min[mid] < a[i]):
ans = mid
low = mid + 1
else :
high = mid - 1
# Print the required answer
print (ans, end = " " )
# Driver code a = [ 3 , 1 , 5 , 2 , 4 ]
n = len (a)
farthest_min(a, n) # This code is contributed by Mohit Kumar |
// C# implementation of the approach using System;
class GFG {
// Function to find the farthest
// smaller number in the right side
static void farthest_min( int [] a, int n)
{
// To store minimum element
// in the range i to n
int [] suffix_min = new int [n];
suffix_min[n - 1] = a[n - 1];
for ( int i = n - 2; i >= 0; i--) {
suffix_min[i]
= Math.Min(suffix_min[i + 1], a[i]);
}
for ( int i = 0; i < n; i++) {
int low = i + 1, high = n - 1, ans = -1;
while (low <= high) {
int mid = (low + high) / 2;
// If current element in the suffix_min
// is less than a[i] then move right
if (suffix_min[mid] < a[i]) {
ans = mid;
low = mid + 1;
}
else
high = mid - 1;
}
// Print the required answer
Console.Write(ans + " " );
}
}
// Driver code
public static void Main()
{
int [] a = { 3, 1, 5, 2, 4 };
int n = a.Length;
farthest_min(a, n);
}
} // This code is contributed by ihritik |
Javascript<script> // Javascript implementation of the approach // Function to find the farthest // smaller number in the right side function farthest_min(a, n)
{ // To store minimum element
// in the range i to n
let suffix_min = new Array(n);
suffix_min[n - 1] = a[n - 1];
for (let i = n - 2; i >= 0; i--) {
suffix_min[i] = Math.min(suffix_min[i + 1], a[i]);
}
for (let i = 0; i < n; i++) {
let low = i + 1, high = n - 1, ans = -1;
while (low <= high) {
let mid = Math.floor((low + high) / 2);
// If current element in the suffix_min
// is less than a[i] then move right
if (suffix_min[mid] < a[i]) {
ans = mid;
low = mid + 1;
}
else
high = mid - 1;
}
// Print the required answer
document.write(ans + " " );
}
} // Driver code let a = [ 3, 1, 5, 2, 4 ];
let n = a.length;
farthest_min(a, n);
//This code is contributed by Mayank Tyagi </script> |
3 -1 4 -1 -1
Time Complexity: O(N* log(N) )
Auxiliary Space: O(N)