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Length of longest subarray with smaller elements on immediate left for each Array element

  • Difficulty Level : Basic
  • Last Updated : 15 Dec, 2021

Given an array arr[] of length N, the task is to find the length of the longest subarray with smaller elements on the immediate left for each element in the given array.

Example:

Input: arr[] = { 2, 1, 7, 6, 7 }
Output: 0 0 2 0 1
Explanation: 
Index 0 (2): No substring is present to the immediate left having all smaller elements. So, length=0
Index 1 (1): No substring is present to the immediate left having all smaller elements. So, length=0
Index 2 (7): Substring {2, 1} is present to the immediate left having all smaller elements. So, length=2
Index 3 (6): No substring is present to the immediate left having all smaller elements. So, length=0
Index 4 (2): Substring {6} is present to the immediate left having all smaller elements. So, length=1

Input: arr[] = { 4, 5, 7, 6, 10 }
Output: 0 1 2 0 4

 

Approach: For every element travel to its left until the element to the left is greater or the array ends to find the length of the longest subarray with smaller elements on the immediate left. Follow the below steps, to solve this problem:

  1. Traverse each element in array arr[].
  2. For each element run another loop in the left direction.
    • Count all the elements smaller than the current element until a greater element comes or the array ends.
  3. Print the answer according to the above observation.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <iostream>
using namespace std;
 
// Function to find the length of the longest
// subarray with smaller elements on the immediate
// left for each element in the given array
void subarrayLength(int* arr, int n)
{
    for (int i = 0; i < n; i++) {
        int ans = 0;
        for (int j = i - 1; j >= 0; j--) {
 
            // If a greater element comes
            if (arr[i] <= arr[j]) {
                break;
            }
            // Else
            ans++;
        }
        cout << ans << " ";
    }
}
 
// Driver Code
int main()
{
 
    int n = 5;
    int arr[n] = { 1, 4, 2, 6, 3 };
    subarrayLength(arr, n);
    return 0;
}

Java




// Java program for the above approach
class GFG{
 
// Function to find the length of the longest
// subarray with smaller elements on the immediate
// left for each element in the given array
static void subarrayLength(int arr[], int n)
{
    for(int i = 0; i < n; i++)
    {
        int ans = 0;
        for(int j = i - 1; j >= 0; j--)
        {
             
            // If a greater element comes
            if (arr[i] <= arr[j])
            {
                break;
            }
             
            // Else
            ans++;
        }
        System.out.print(ans + " ");
    }
}
 
// Driver Code
public static void main(String args[])
{
    int n = 5;
    int arr[] = { 1, 4, 2, 6, 3 };
     
    subarrayLength(arr, n);
}
}
 
// This code is contributed by gfgking

Python3




# Python program for the above approach
 
# Function to find the length of the longest
# subarray with smaller elements on the immediate
# left for each element in the given array
def subarrayLength(arr, n):
 
    for i in range(0, n):
        ans = 0
        for j in range(i-1, -1, -1):
 
            # If a greater element comes
            if (arr[i] <= arr[j]):
                break
 
            # Else
            ans += 1
        print(ans, end=" ")
 
# Driver Code
if __name__ == "__main__":
 
    n = 5
    arr = [1, 4, 2, 6, 3]
    subarrayLength(arr, n)
 
# This code is contributed by rakeshsahni

C#




// C# program for the above approach
using System;
class GFG{
 
// Function to find the length of the longest
// subarray with smaller elements on the immediate
// left for each element in the given array
static void subarrayLength(int []arr, int n)
{
    for(int i = 0; i < n; i++)
    {
        int ans = 0;
        for(int j = i - 1; j >= 0; j--)
        {
             
            // If a greater element comes
            if (arr[i] <= arr[j])
            {
                break;
            }
             
            // Else
            ans++;
        }
        Console.Write(ans + " ");
    }
}
 
// Driver Code
public static void Main()
{
    int n = 5;
    int []arr = { 1, 4, 2, 6, 3 };
     
    subarrayLength(arr, n);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
      // JavaScript code for the above approach
 
      // Function to find the length of the longest
      // subarray with smaller elements on the immediate
      // left for each element in the given array
      function subarrayLength(arr, n)
      {
          for (let i = 0; i < n; i++)
          {
              let ans = 0;
              for (let j = i - 1; j >= 0; j--) {
 
                  // If a greater element comes
                  if (arr[i] <= arr[j]) {
                      break;
                  }
                   
                  // Else
                  ans++;
              }
              document.write(ans + " ");
          }
      }
 
      // Driver Code
      let n = 5;
      let arr = [1, 4, 2, 6, 3];
      subarrayLength(arr, n);
 
// This code is contributed by Potta Lokesh
  </script>
Output
0 1 0 3 0 

Time Complexity: O(N2)
Auxiliary Space: O(1)

 


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