# Find the farthest smaller number in the right side

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

Recommended Practice

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++

 `// C++ implementation of the approach``#include ``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

 `// 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`

## Python3

 `# 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#

 `// 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.`

## Javascript

 `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.`

Output
`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++

 `// C++ implementation of the approach``#include ``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

 `// 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

 `# 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#

 `// 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

 `Javascript`

Output
`3 -1 4 -1 -1 `

Time Complexity: O(N* log(N) )
Auxiliary Space: O(N)

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