# Largest element smaller than current element on left for every element in Array

Given an array **arr[]** of the positive integers of size **N**, the task is to find the largest element on the left side of each index which is smaller than the element present at that index.

**Note:** If no such element is found then print **-1**.

**Examples:**

Input:arr[] = {2, 5, 10}

Output:-1 2 5

Explanation :

Index 0:There are no elements before it

So Print -1 for the index 0

Index 1:Elements less than before index 1 are – {2}

Maximum of those elements is 2

Index 2:Elements less than before index 2 are – {2, 5}

Maximum of those elements is 5

Input:arr[] = {4, 7, 6, 8, 5}

Output:-1 4 4 7 4

Explanation :

Index 0:There are no elements before it

So Print -1 for the index 0

Index 1:Elements less than before index 1 are – {4}

Maximum of those elements is 4

Index 2:Elements less than before index 2 are – {4}

Maximum of those elements is 4

Index 3:Elements less than before index 3 are – {4, 7, 6}

Maximum of those elements is 7

Index 4:Elements less than before index 4 are – {4}

Maximum of those elements is 4

**Naive Approach:** A simple solution is to use two nested loops.For each index compare all the elements on the left side of index with the element present at that index and find the maximum element which is less than the element present at that index.

**Algorithm:**

- Run a loop with a loop variable
**i**from 0 to length – 1, where length is the length of the array.- For every element Intialize
**maximum_till_now**to -1 because maximum will always be greater than -1, If there exists a smaller element. - Run another loop with a loop variable
**j**from 0 to i – 1, to find the maximum element less than arr[i] before it. - Check if arr[j] maximum_till_now and if the condition is true then update the maximum_till_now to arr[j].

- For every element Intialize
- The variable
**maximum_till_now**will have the maximum element before it which is less than arr[i].

## CPP

`// C++ implementation to find the ` `// Largest element before every element ` `// of an array such that ` `// it is less than the element ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the ` `// Largest element before ` `// every element of an array ` `void` `findMaximumBefore(` `int` `arr[], ` ` ` `int` `n){ ` ` ` ` ` `// Loop to iterate over every ` ` ` `// element of the array ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `int` `currAns = -1; ` ` ` ` ` `// Loop to find the maximum smallest ` ` ` `// number before the element arr[i] ` ` ` `for` `(` `int` `j = i - 1; j >= 0; j--) { ` ` ` `if` `(arr[j] > currAns && ` ` ` `arr[j] < arr[i]) { ` ` ` `currAns = arr[j]; ` ` ` `} ` ` ` `} ` ` ` `cout << currAns << ` `" "` `; ` ` ` `} ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 4, 7, 6, 8, 5 }; ` ` ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` ` ` `// Function Call ` ` ` `findMaximumBefore(arr, n); ` `} ` |

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## Java

`// Java implementation to find the ` `// Largest element before every element ` `// of an array such that ` `// it is less than the element ` `import` `java.util.*; ` ` ` `class` `GFG{ ` ` ` `// Function to find the ` `// Largest element before ` `// every element of an array ` `static` `void` `findMaximumBefore(` `int` `arr[], ` ` ` `int` `n){ ` ` ` ` ` `// Loop to iterate over every ` ` ` `// element of the array ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) { ` ` ` ` ` `int` `currAns = -` `1` `; ` ` ` ` ` `// Loop to find the maximum smallest ` ` ` `// number before the element arr[i] ` ` ` `for` `(` `int` `j = i - ` `1` `; j >= ` `0` `; j--) { ` ` ` `if` `(arr[j] > currAns && ` ` ` `arr[j] < arr[i]) { ` ` ` `currAns = arr[j]; ` ` ` `} ` ` ` `} ` ` ` `System.out.print(currAns+ ` `" "` `); ` ` ` `} ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `arr[] = { ` `4` `, ` `7` `, ` `6` `, ` `8` `, ` `5` `}; ` ` ` ` ` `int` `n = arr.length; ` ` ` ` ` `// Function Call ` ` ` `findMaximumBefore(arr, n); ` `} ` `} ` ` ` `// This code is contributed by 29AjayKumar ` |

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## Python3

`# Python3 implementation to find the ` `# Largest element before every element ` `# of an array such that ` `# it is less than the element ` ` ` `# Function to find the ` `# Largest element before ` `# every element of an array ` `def` `findMaximumBefore(arr, n): ` ` ` ` ` `# Loop to iterate over every ` ` ` `# element of the array ` ` ` `for` `i ` `in` `range` `(n): ` ` ` ` ` `currAns ` `=` `-` `1` ` ` ` ` `# Loop to find the maximum smallest ` ` ` `# number before the element arr[i] ` ` ` `for` `j ` `in` `range` `(i` `-` `1` `,` `-` `1` `,` `-` `1` `): ` ` ` `if` `(arr[j] > currAns ` `and` ` ` `arr[j] < arr[i]): ` ` ` `currAns ` `=` `arr[j] ` ` ` ` ` `print` `(currAns,end` `=` `" "` `) ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` ` ` `arr` `=` `[` `4` `, ` `7` `, ` `6` `, ` `8` `, ` `5` `] ` ` ` ` ` `n ` `=` `len` `(arr) ` ` ` ` ` `# Function Call ` ` ` `findMaximumBefore(arr, n) ` ` ` `# This code is contributed by mohit kumar 29 ` |

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## C#

`// C# implementation to find the ` `// Largest element before every element ` `// of an array such that ` `// it is less than the element ` `using` `System; ` ` ` `class` `GFG{ ` ` ` `// Function to find the ` `// Largest element before ` `// every element of an array ` `static` `void` `findMaximumBefore(` `int` `[]arr, ` ` ` `int` `n){ ` ` ` ` ` `// Loop to iterate over every ` ` ` `// element of the array ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `int` `currAns = -1; ` ` ` ` ` `// Loop to find the maximum smallest ` ` ` `// number before the element arr[i] ` ` ` `for` `(` `int` `j = i - 1; j >= 0; j--) { ` ` ` `if` `(arr[j] > currAns && ` ` ` `arr[j] < arr[i]) { ` ` ` `currAns = arr[j]; ` ` ` `} ` ` ` `} ` ` ` `Console.Write(currAns+ ` `" "` `); ` ` ` `} ` `} ` ` ` `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` `int` `[]arr = { 4, 7, 6, 8, 5 }; ` ` ` ` ` `int` `n = arr.Length; ` ` ` ` ` `// Function Call ` ` ` `findMaximumBefore(arr, n); ` `} ` `} ` ` ` `// This code is contributed by Rajput-Ji ` |

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**Output:**

-1 4 4 7 4

**Performance Analysis:**

**Time Complexity:**O(N^{2}).**Auxiliary Space:**O(1).

**Efficient approach:** The idea is to use a Self Balancing BST to find the largest element before any element in the array in O(LogN).

Self Balancing BST is implemented as set in C++ and Treeset in Java.

**Algorithm:**

- Declare a Self Balancing BST to store the elements of the array.
- Iterate over the array with a loop variable
**i**from 0 to length – 1.- Insert the element in the Self Balancing BST in O(LogN) time.
- Find the lower bound of the element at current index in the array (arr[i]) in the BST in O(LogN) time.

Below is the implementation of the above approach:

`// C++ implementation to find the ` `// Largest element before every element ` `// of an array such that ` `// it is less than the element ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the ` `// Largest element before ` `// every element of an array ` `void` `findMaximumBefore(` `int` `arr[], ` ` ` `int` `n){ ` ` ` `// Self Balancing BST ` ` ` `set<` `int` `> s; ` ` ` `set<` `int` `>::iterator it; ` ` ` ` ` `// Loop to iterate over the ` ` ` `// elements of the array ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// Insertion in BST ` ` ` `s.insert(arr[i]); ` ` ` ` ` `// Lower Bound the element arr[i] ` ` ` `it = s.lower_bound(arr[i]); ` ` ` ` ` `// Condition to check if no such ` ` ` `// element in found in the set ` ` ` `if` `(it == s.begin()) { ` ` ` `cout << ` `"-1"` ` ` `<< ` `" "` `; ` ` ` `} ` ` ` `else` `{ ` ` ` `it--; ` ` ` `cout << (*it) << ` `" "` `; ` ` ` `} ` ` ` `} ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 4, 7, 6, 8, 5 }; ` ` ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` ` ` `findMaximumBefore(arr, n); ` `} ` |

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**Output:**

-1 4 4 7 4

**Performance Analysis:**

**Time Complexity:**O(NlogN).**Auxiliary Space:**O(N).

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