Given a sorted array arr[] and a number x, write a function that counts the occurrences of x in arr[]. Expected time complexity is O(Logn)
Examples:
Input: arr[] = {1, 1, 2, 2, 2, 2, 3,}, x = 2 Output: 4 // x (or 2) occurs 4 times in arr[] Input: arr[] = {1, 1, 2, 2, 2, 2, 3,}, x = 3 Output: 1 Input: arr[] = {1, 1, 2, 2, 2, 2, 3,}, x = 1 Output: 2 Input: arr[] = {1, 1, 2, 2, 2, 2, 3,}, x = 4 Output: -1 // 4 doesn't occur in arr[]
Method 1 (Linear Search)
Linearly search for x, count the occurrences of x and return the count.
C++
// C++ program to count occurrences of an element #include<bits/stdc++.h> using namespace std; // Returns number of times x occurs in arr[0..n-1] int countOccurrences( int arr[], int n, int x) { int res = 0; for ( int i=0; i<n; i++) if (x == arr[i]) res++; return res; } // Driver code int main() { int arr[] = {1, 2, 2, 2, 2, 3, 4, 7 ,8 ,8 }; int n = sizeof (arr)/ sizeof (arr[0]); int x = 2; cout << countOccurrences(arr, n, x); return 0; } |
Java
// Java program to count occurrences // of an element class Main { // Returns number of times x occurs in arr[0..n-1] static int countOccurrences( int arr[], int n, int x) { int res = 0 ; for ( int i= 0 ; i<n; i++) if (x == arr[i]) res++; return res; } public static void main(String args[]) { int arr[] = { 1 , 2 , 2 , 2 , 2 , 3 , 4 , 7 , 8 , 8 }; int n = arr.length; int x = 2 ; System.out.println(countOccurrences(arr, n, x)); } } |
Python3
# Python3 program to count # occurrences of an element # Returns number of times x # occurs in arr[0..n-1] def countOccurrences(arr, n, x): res = 0 for i in range (n): if x = = arr[i]: res + = 1 return res # Driver code arr = [ 1 , 2 , 2 , 2 , 2 , 3 , 4 , 7 , 8 , 8 ] n = len (arr) x = 2 print (countOccurrences(arr, n, x)) |
C#
// C# program to count occurrences // of an element using System; class GFG { // Returns number of times x // occurs in arr[0..n-1] static int countOccurrences( int []arr, int n, int x) { int res = 0; for ( int i = 0; i < n; i++) if (x == arr[i]) res++; return res; } // driver code public static void Main() { int []arr = {1, 2, 2, 2, 2, 3, 4, 7 ,8 ,8 }; int n = arr.Length; int x = 2; Console.Write(countOccurrences(arr, n, x)); } } // This code is contributed by Sam007 |
PHP
<?php // PHP program to count occurrences // of an element // Returns number of times x // occurs in arr[0..n-1] function countOccurrences( $arr , $n , $x ) { $res = 0; for ( $i = 0; $i < $n ; $i ++) if ( $x == $arr [ $i ]) $res ++; return $res ; } // Driver code $arr = array (1, 2, 2, 2, 2, 3, 4, 7 ,8 ,8 ); $n = count ( $arr ); $x = 2; echo countOccurrences( $arr , $n , $x ); // This code is contributed by Sam007 ?> |
Output :
4
Time Complexity: O(n)
Method 2 (Better using Binary Search)
We first find an occurrence using binary search. Then we match toward left and right sides of the matched the found index.
C++
// C++ program to count occurrences of an element #include <bits/stdc++.h> using namespace std; // A recursive binary search function. It returns // location of x in given array arr[l..r] is present, // otherwise -1 int binarySearch( int arr[], int l, int r, int x) { if (r < l) return -1; int mid = l + (r - l) / 2; // If the element is present at the middle // itself if (arr[mid] == x) return mid; // If element is smaller than mid, then // it can only be present in left subarray if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x); // Else the element can only be present // in right subarray return binarySearch(arr, mid + 1, r, x); } // Returns number of times x occurs in arr[0..n-1] int countOccurrences( int arr[], int n, int x) { int ind = binarySearch(arr, 0, n - 1, x); // If element is not present if (ind == -1) return 0; // Count elements on left side. int count = 1; int left = ind - 1; while (left >= 0 && arr[left] == x) count++, left--; // Count elements on right side. int right = ind + 1; while (right < n && arr[right] == x) count++, right++; return count; } // Driver code int main() { int arr[] = { 1, 2, 2, 2, 2, 3, 4, 7, 8, 8 }; int n = sizeof (arr) / sizeof (arr[0]); int x = 2; cout << countOccurrences(arr, n, x); return 0; } |
Java
// Java program to count // occurrences of an element class GFG { // A recursive binary search // function. It returns location // of x in given array arr[l..r] // is present, otherwise -1 static int binarySearch( int arr[], int l, int r, int x) { if (r < l) return - 1 ; int mid = l + (r - l) / 2 ; // If the element is present // at the middle itself if (arr[mid] == x) return mid; // If element is smaller than // mid, then it can only be // present in left subarray if (arr[mid] > x) return binarySearch(arr, l, mid - 1 , x); // Else the element can // only be present in // right subarray return binarySearch(arr, mid + 1 , r, x); } // Returns number of times x // occurs in arr[0..n-1] static int countOccurrences( int arr[], int n, int x) { int ind = binarySearch(arr, 0 , n - 1 , x); // If element is not present if (ind == - 1 ) return 0 ; // Count elements on left side. int count = 1 ; int left = ind - 1 ; while (left >= 0 && arr[left] == x) { count++; left--; } // Count elements // on right side. int right = ind + 1 ; while (right < n && arr[right] == x) { count++; right++; } return count; } // Driver code public static void main(String[] args) { int arr[] = { 1 , 2 , 2 , 2 , 2 , 3 , 4 , 7 , 8 , 8 }; int n = arr.length; int x = 2 ; System.out.print(countOccurrences(arr, n, x)); } } // This code is contributed // by ChitraNayal |
Python 3
# Python 3 program to count # occurrences of an element # A recursive binary search # function. It returns location # of x in given array arr[l..r] # is present, otherwise -1 def binarySearch(arr, l, r, x): if (r < l): return - 1 mid = int ( l + (r - l) / 2 ) # If the element is present # at the middle itself if arr[mid] = = x: return mid # If element is smaller than # mid, then it can only be # present in left subarray if arr[mid] > x: return binarySearch(arr, l, mid - 1 , x) # Else the element # can only be present # in right subarray return binarySearch(arr, mid + 1 , r, x) # Returns number of times # x occurs in arr[0..n-1] def countOccurrences(arr, n, x): ind = binarySearch(arr, 0 , n - 1 , x) # If element is not present if ind = = - 1 : return 0 # Count elements # on left side. count = 1 left = ind - 1 while (left > = 0 and arr[left] = = x): count + = 1 left - = 1 # Count elements on # right side. right = ind + 1 ; while (right < n and arr[right] = = x): count + = 1 right + = 1 return count # Driver code arr = [ 1 , 2 , 2 , 2 , 2 , 3 , 4 , 7 , 8 , 8 ] n = len (arr) x = 2 print (countOccurrences(arr, n, x)) # This code is contributed # by ChitraNayal |
C#
// C# program to count // occurrences of an element using System; class GFG { // A recursive binary search // function. It returns location // of x in given array arr[l..r] // is present, otherwise -1 static int binarySearch( int [] arr, int l, int r, int x) { if (r < l) return -1; int mid = l + (r - l) / 2; // If the element is present // at the middle itself if (arr[mid] == x) return mid; // If element is smaller than // mid, then it can only be // present in left subarray if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x); // Else the element // can only be present // in right subarray return binarySearch(arr, mid + 1, r, x); } // Returns number of times x // occurs in arr[0..n-1] static int countOccurrences( int [] arr, int n, int x) { int ind = binarySearch(arr, 0, n - 1, x); // If element is not present if (ind == -1) return 0; // Count elements on left side. int count = 1; int left = ind - 1; while (left >= 0 && arr[left] == x) { count++; left--; } // Count elements on right side. int right = ind + 1; while (right < n && arr[right] == x) { count++; right++; } return count; } // Driver code public static void Main() { int [] arr = {1, 2, 2, 2, 2, 3, 4, 7, 8, 8}; int n = arr.Length; int x = 2; Console.Write(countOccurrences(arr, n, x)); } } // This code is contributed // by ChitraNayal |
PHP
<?php // PHP program to count // occurrences of an element // A recursive binary search // function. It returns location // of x in given array arr[l..r] // is present, otherwise -1 function binarySearch(& $arr , $l , $r , $x ) { if ( $r < $l ) return -1; $mid = $l + ( $r - $l ) / 2; // If the element is present // at the middle itself if ( $arr [ $mid ] == $x ) return $mid ; // If element is smaller than // mid, then it can only be // present in left subarray if ( $arr [ $mid ] > $x ) return binarySearch( $arr , $l , $mid - 1, $x ); // Else the element // can only be present // in right subarray return binarySearch( $arr , $mid + 1, $r , $x ); } // Returns number of times // x occurs in arr[0..n-1] function countOccurrences( $arr , $n , $x ) { $ind = binarySearch( $arr , 0, $n - 1, $x ); // If element is not present if ( $ind == -1) return 0; // Count elements // on left side. $count = 1; $left = $ind - 1; while ( $left >= 0 && $arr [ $left ] == $x ) { $count ++; $left --; } // Count elements on right side. $right = $ind + 1; while ( $right < $n && $arr [ $right ] == $x ) { $count ++; $right ++; } return $count ; } // Driver code $arr = array ( 1, 2, 2, 2, 2, 3, 4, 7, 8, 8 ); $n = sizeof( $arr ); $x = 2; echo countOccurrences( $arr , $n , $x ); // This code is contributed // by ChitraNayal ?> |
Output :
4
Time Complexity : O(Log n + count) where count is number of occurrences.
Method 3 (Best using Improved Binary Search)
1) Use Binary search to get index of the first occurrence of x in arr[]. Let the index of the first occurrence be i.
2) Use Binary search to get index of the last occurrence of x in arr[]. Let the index of the last occurrence be j.
3) Return (j – i + 1);
C++
// C++ program to count occurrences of an element // in a sorted array. # include <bits/stdc++.h> using namespace std; /* if x is present in arr[] then returns the count of occurrences of x, otherwise returns 0. */ int count( int arr[], int x, int n) { /* get the index of first occurrence of x */ int *low = lower_bound(arr, arr+n, x); // If element is not present, return 0 if (low == (arr + n) || *low != x) return 0; /* Else get the index of last occurrence of x. Note that we are only looking in the subarray after first occurrence */ int *high = upper_bound(low, arr+n, x); /* return count */ return high - low; } /* driver program to test above functions */ int main() { int arr[] = {1, 2, 2, 3, 3, 3, 3}; int x = 3; // Element to be counted in arr[] int n = sizeof (arr)/ sizeof (arr[0]); int c = count(arr, x, n); printf ( " %d occurs %d times " , x, c); return 0; } |
C
# include <stdio.h> /* if x is present in arr[] then returns the index of FIRST occurrence of x in arr[0..n-1], otherwise returns -1 */ int first( int arr[], int low, int high, int x, int n) { if (high >= low) { int mid = (low + high)/2; /*low + (high - low)/2;*/ if ( ( mid == 0 || x > arr[mid-1]) && arr[mid] == x) return mid; else if (x > arr[mid]) return first(arr, (mid + 1), high, x, n); else return first(arr, low, (mid -1), x, n); } return -1; } /* if x is present in arr[] then returns the index of LAST occurrence of x in arr[0..n-1], otherwise returns -1 */ int last( int arr[], int low, int high, int x, int n) { if (high >= low) { int mid = (low + high)/2; /*low + (high - low)/2;*/ if ( ( mid == n-1 || x < arr[mid+1]) && arr[mid] == x ) return mid; else if (x < arr[mid]) return last(arr, low, (mid -1), x, n); else return last(arr, (mid + 1), high, x, n); } return -1; } /* if x is present in arr[] then returns the count of occurrences of x, otherwise returns -1. */ int count( int arr[], int x, int n) { int i; // index of first occurrence of x in arr[0..n-1] int j; // index of last occurrence of x in arr[0..n-1] /* get the index of first occurrence of x */ i = first(arr, 0, n-1, x, n); /* If x doesn't exist in arr[] then return -1 */ if (i == -1) return i; /* Else get the index of last occurrence of x. Note that we are only looking in the subarray after first occurrence */ j = last(arr, i, n-1, x, n); /* return count */ return j-i+1; } /* driver program to test above functions */ int main() { int arr[] = {1, 2, 2, 3, 3, 3, 3}; int x = 3; // Element to be counted in arr[] int n = sizeof (arr)/ sizeof (arr[0]); int c = count(arr, x, n); printf ( " %d occurs %d times " , x, c); getchar (); return 0; } |
Java
// Java program to count occurrences // of an element class Main { /* if x is present in arr[] then returns the count of occurrences of x, otherwise returns -1. */ static int count( int arr[], int x, int n) { // index of first occurrence of x in arr[0..n-1] int i; // index of last occurrence of x in arr[0..n-1] int j; /* get the index of first occurrence of x */ i = first(arr, 0 , n- 1 , x, n); /* If x doesn't exist in arr[] then return -1 */ if (i == - 1 ) return i; /* Else get the index of last occurrence of x. Note that we are only looking in the subarray after first occurrence */ j = last(arr, i, n- 1 , x, n); /* return count */ return j-i+ 1 ; } /* if x is present in arr[] then returns the index of FIRST occurrence of x in arr[0..n-1], otherwise returns -1 */ static int first( int arr[], int low, int high, int x, int n) { if (high >= low) { /*low + (high - low)/2;*/ int mid = (low + high)/ 2 ; if ( ( mid == 0 || x > arr[mid- 1 ]) && arr[mid] == x) return mid; else if (x > arr[mid]) return first(arr, (mid + 1 ), high, x, n); else return first(arr, low, (mid - 1 ), x, n); } return - 1 ; } /* if x is present in arr[] then returns the index of LAST occurrence of x in arr[0..n-1], otherwise returns -1 */ static int last( int arr[], int low, int high, int x, int n) { if (high >= low) { /*low + (high - low)/2;*/ int mid = (low + high)/ 2 ; if ( ( mid == n- 1 || x < arr[mid+ 1 ]) && arr[mid] == x ) return mid; else if (x < arr[mid]) return last(arr, low, (mid - 1 ), x, n); else return last(arr, (mid + 1 ), high, x, n); } return - 1 ; } public static void main(String args[]) { int arr[] = { 1 , 2 , 2 , 3 , 3 , 3 , 3 }; // Element to be counted in arr[] int x = 3 ; int n = arr.length; int c = count(arr, x, n); System.out.println(x+ " occurs " +c+ " times" ); } } |
Python3
# Python3 program to count # occurrences of an element # if x is present in arr[] then # returns the count of occurrences # of x, otherwise returns -1. def count(arr, x, n): # get the index of first # occurrence of x i = first(arr, 0 , n - 1 , x, n) # If x doesn't exist in # arr[] then return -1 if i = = - 1 : return i # Else get the index of last occurrence # of x. Note that we are only looking # in the subarray after first occurrence j = last(arr, i, n - 1 , x, n); # return count return j - i + 1 ; # if x is present in arr[] then return # the index of FIRST occurrence of x in # arr[0..n-1], otherwise returns -1 def first(arr, low, high, x, n): if high > = low: # low + (high - low)/2 mid = (low + high) / / 2 if (mid = = 0 or x > arr[mid - 1 ]) and arr[mid] = = x: return mid elif x > arr[mid]: return first(arr, (mid + 1 ), high, x, n) else : return first(arr, low, (mid - 1 ), x, n) return - 1 ; # if x is present in arr[] then return # the index of LAST occurrence of x # in arr[0..n-1], otherwise returns -1 def last(arr, low, high, x, n): if high > = low: # low + (high - low)/2 mid = (low + high) / / 2 ; if (mid = = n - 1 or x < arr[mid + 1 ]) and arr[mid] = = x : return mid elif x < arr[mid]: return last(arr, low, (mid - 1 ), x, n) else : return last(arr, (mid + 1 ), high, x, n) return - 1 # driver program to test above functions arr = [ 1 , 2 , 2 , 3 , 3 , 3 , 3 ] x = 3 # Element to be counted in arr[] n = len (arr) c = count(arr, x, n) print ( "%d occurs %d times " % (x, c)) |
C#
// C# program to count occurrences // of an element using System; class GFG { /* if x is present in arr[] then returns the count of occurrences of x, otherwise returns -1. */ static int count( int []arr, int x, int n) { // index of first occurrence of x in arr[0..n-1] int i; // index of last occurrence of x in arr[0..n-1] int j; /* get the index of first occurrence of x */ i = first(arr, 0, n-1, x, n); /* If x doesn't exist in arr[] then return -1 */ if (i == -1) return i; /* Else get the index of last occurrence of x. Note that we are only looking in the subarray after first occurrence */ j = last(arr, i, n-1, x, n); /* return count */ return j-i+1; } /* if x is present in arr[] then returns the index of FIRST occurrence of x in arr[0..n-1], otherwise returns -1 */ static int first( int []arr, int low, int high, int x, int n) { if (high >= low) { /*low + (high - low)/2;*/ int mid = (low + high)/2; if ( ( mid == 0 || x > arr[mid-1]) && arr[mid] == x) return mid; else if (x > arr[mid]) return first(arr, (mid + 1), high, x, n); else return first(arr, low, (mid -1), x, n); } return -1; } /* if x is present in arr[] then returns the index of LAST occurrence of x in arr[0..n-1], otherwise returns -1 */ static int last( int []arr, int low, int high, int x, int n) { if (high >= low) { /*low + (high - low)/2;*/ int mid = (low + high)/2; if ( ( mid == n-1 || x < arr[mid+1]) && arr[mid] == x ) return mid; else if (x < arr[mid]) return last(arr, low, (mid -1), x, n); else return last(arr, (mid + 1), high, x, n); } return -1; } public static void Main() { int []arr = {1, 2, 2, 3, 3, 3, 3}; // Element to be counted in arr[] int x = 3; int n = arr.Length; int c = count(arr, x, n); Console.Write(x + " occurs " + c + " times" ); } } // This code is contributed by Sam007 |
Output:
3 occurs 4 times
Time Complexity: O(Logn)
Programming Paradigm: Divide & Conquer
Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
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