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# Lower Insertion Point

• Last Updated : 16 Mar, 2021

Given an array arr[] of n sorted integer elements and an integer X, the task is to find the lower insertion point of X in the array. The lower insertion point is the index of the first element that is ≥ X. If X is greater than all the elements of arr then print n and if X is less than all the elements of arr[] then return 0.
Examples:

Input: arr[] = {2, 3, 4, 4, 5, 6, 7, 9}, X = 4
Output: 2
Input: arr[] = {0, 5, 8, 15}, X = 16
Output:

Approach:

• If X < arr[0] print 0 or X > arr[n – 1] print n.
• Initialise lowertPnt = 0 and start traversing the array from 1 to n – 1
• If arr[i] < X then update lowerPnt = i and i = i * 2.
• The first value of i for which X ≥ arr[i] or when i ≥ n, break out of the loop.
• Now check for the rest of the elements from lowerPnt to n – 1, while arr[lowerPnt] < X update lowerPnt = lowerPnt + 1.
• Print lowerPnt in the end..

Below is the implementation of the above approach:

## C++

 // C++ program to find the lower insertion point// of an element in a sorted array#include using namespace std; // Function to return the lower insertion point// of an element in a sorted arrayint LowerInsertionPoint(int arr[], int n, int X){     // Base cases    if (X < arr[0])        return 0;    else if (X > arr[n - 1])        return n;     int lowerPnt = 0;    int i = 1;     while (i < n && arr[i] < X) {        lowerPnt = i;        i = i * 2;    }     // Final check for the remaining elements which are < X    while (lowerPnt < n && arr[lowerPnt] < X)        lowerPnt++;     return lowerPnt;} // Driver codeint main(){    int arr[] = { 2, 3, 4, 4, 5, 6, 7, 9 };    int n = sizeof(arr) / sizeof(arr[0]);    int X = 4;    cout << LowerInsertionPoint(arr, n, X);    return 0;}

## Java

 //Java program to find the lower insertion point//of an element in a sorted arraypublic class AQES {     //Function to return the lower insertion point    //of an element in a sorted array    static int LowerInsertionPoint(int arr[], int n, int X)    {      // Base cases     if (X < arr[0])         return 0;     else if (X > arr[n - 1])         return n;      int lowerPnt = 0;     int i = 1;      while (i < n && arr[i] < X) {         lowerPnt = i;         i = i * 2;     }      // Final check for the remaining elements which are < X     while (lowerPnt < n && arr[lowerPnt] < X)         lowerPnt++;      return lowerPnt;    }     //Driver code    public static void main(String[] args) {                  int arr[] = { 2, 3, 4, 4, 5, 6, 7, 9 };         int n = arr.length;         int X = 4;         System.out.println(LowerInsertionPoint(arr, n, X));     }}

## Python3

 # Python3 program to find the lower insertion# point of an element in a sorted array # Function to return the lower insertion# point of an element in a sorted arraydef LowerInsertionPoint(arr, n, X) :     # Base cases    if (X < arr[0]) :        return 0;    elif (X > arr[n - 1]) :        return n     lowerPnt = 0    i = 1     while (i < n and arr[i] < X) :        lowerPnt = i        i = i * 2     # Final check for the remaining elements    # which are < X    while (lowerPnt < n and arr[lowerPnt] < X) :        lowerPnt += 1     return lowerPnt # Driver codeif __name__ == "__main__" :     arr = [ 2, 3, 4, 4, 5, 6, 7, 9 ]    n = len(arr)    X = 4    print(LowerInsertionPoint(arr, n, X)) # This code is contributed by Ryuga

## C#

 // C#  program to find the lower insertion point//of an element in a sorted arrayusing System; public class GFG{    //Function to return the lower insertion point    //of an element in a sorted array    static int LowerInsertionPoint(int []arr, int n, int X)    {     // Base cases    if (X < arr[0])        return 0;    else if (X > arr[n - 1])        return n;     int lowerPnt = 0;    int i = 1;     while (i < n && arr[i] < X) {        lowerPnt = i;        i = i * 2;    }     // Final check for the remaining elements which are < X    while (lowerPnt < n && arr[lowerPnt] < X)        lowerPnt++;     return lowerPnt;    }     //Driver code    static public void Main (){        int []arr = { 2, 3, 4, 4, 5, 6, 7, 9 };        int n = arr.Length;        int X = 4;        Console.WriteLine(LowerInsertionPoint(arr, n, X));    }}

## PHP

 \$arr[\$n - 1])        return \$n;     \$lowerPnt = 0;    \$i = 1;     while (\$i < \$n && \$arr[\$i] < \$X)    {        \$lowerPnt = \$i;        \$i = \$i * 2;    }     // Final check for the remaining    // elements which are < X    while (\$lowerPnt < \$n && \$arr[\$lowerPnt] < \$X)        \$lowerPnt++;     return \$lowerPnt;} // Driver code\$arr = array( 2, 3, 4, 4, 5, 6, 7, 9 );\$n = sizeof(\$arr);\$X = 4;echo LowerInsertionPoint(\$arr, \$n, \$X); // This code is contributed by ajit.?>

## Javascript


Output:
2

Further Optimization : The time complexity of the above solution can become O(n) in worst case. We can optimize the solution to work in O(Log n) time using Binary Search. Please refer unbounded binary search for details.

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