Count pairs in an array that hold i*arr[i] > j*arr[j]

Given an array of integers arr[0..n-1], count all pairs (arr[i], arr[j]) in the such that i*arr[i] > j*arr[j], 0 =< i < j < n.

Examples :

Input : arr[] = {5 , 0, 10, 2, 4, 1, 6}
Output: 5
Pairs which hold condition i*arr[i] > j*arr[j]
are (10, 2) (10, 4) (10, 1) (2, 1) (4, 1)

Input  : arr[] = {8, 4, 2, 1}
Output : 2

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A Simple solution is to run two loops. Pick each element of array one-by-one and for each element find element on right side of array that hold condition, then increment counter and last return counter value.
Below is the implementation of above idea:

C++

 // C++ program to count all pair that // hold condition i*arr[i] > j*arr[j] #include using namespace std;    // Return count of pair in given array // such that  i*arr[i] > j*arr[j] int CountPair(int arr[] , int n ) {     int result = 0; // Initialize result        for (int i=0; i j*arr[j] )                 result ++;     }     return result; }    // Driver code int main() {     int arr[] = {5 , 0, 10, 2, 4, 1, 6} ;     int n = sizeof(arr)/sizeof(arr);     cout << "Count of Pairs : "          << CountPair(arr, n);     return 0; }

Java

 // Java Code for Count pairs in an // array that hold i*arr[i] > j*arr[j] class GFG {            // Return count of pair in given array     // such that  i*arr[i] > j*arr[j]     public static int CountPair(int arr[] , int n )     {         int result = 0; // Initialize result                 for (int i = 0; i < n; i++)         {             // Generate all pair and increment             // counter if the hold given condition             for (int j = i + 1; j < n; j++)                 if (i*arr[i] > j*arr[j] )                     result ++;         }         return result;     }            /* Driver program to test above function */     public static void main(String[] args)      {         int arr[] = {5 , 0, 10, 2, 4, 1, 6} ;         int n = arr.length;         System.out.println("Count of Pairs : " +                             CountPair(arr, n));     }   } // This code is contributed by Arnav Kr. Mandal.

Python3

 # C# Code to Count pairs in an # array that hold i*arr[i] > j*arr[j]    # Return count of pair in given array # such that i*arr[i] > j*arr[j] def CountPair(arr , n ):            # Initialize result     result = 0;            for i in range (0, n):                    # Generate all pair and increment         # counter if the hold given condition         j = i + 1         while(j < n):             if (i * arr[i] > j * arr[j] ):                 result = result +1             j = j + 1     return result;        # Driver program to test above function */        arr = [5, 0, 10, 2, 4, 1, 6] n = len(arr) print("Count of Pairs : " , CountPair(arr, n))    # This code is contributed by Sam007.

C#

 // C# Code to Count pairs in an // array that hold i*arr[i] > j*arr[j] using System;    class GFG {     // Return count of pair in given array     // such that i*arr[i] > j*arr[j]     public static int CountPair(int []arr , int n )     {         // Initialize result         int result = 0;                for (int i = 0; i < n; i++)         {             // Generate all pair and increment             // counter if the hold given condition             for (int j = i + 1; j < n; j++)                 if (i*arr[i] > j*arr[j] )                     result ++;         }         return result;     }            /* Driver program to test above function */     public static void Main()      {         int []arr = {5, 0, 10, 2, 4, 1, 6};         int n = arr.Length;         Console.WriteLine("Count of Pairs : " +                            CountPair(arr, n));     } }    // This code is contributed by Sam007

PHP

 j*arr[j]    // Return count of pair in given array // such that i*arr[i] > j*arr[j] function CountPair(\$arr , \$n ) {            // Initialize result     \$result = 0;                for(\$i = 0; \$i < \$n; \$i++)     {                    // Generate all pair and increment         // counter if the hold given condition         for (\$j = \$i + 1; \$j < \$n; \$j++)             if (\$i *  \$arr[\$i] > \$j * \$arr[\$j] )                 \$result ++;     }     return \$result; }        // Driver code     \$arr = array(5, 0, 10, 2, 4, 1, 6) ;     \$n = sizeof(\$arr);     echo "Count of Pairs : ",     CountPair(\$arr, \$n);        // This code is contributed by m_kit ?>

Output:

Count of Pairs : 5

Time Complexity: O(n2)

An efficient solution of this problem takes O(n log n) time. The idea is based on an interesting fact about this problem that after modifying the array such that every element is multiplied with its index, this problem convert into Count Inversions in an array.
Algorithm :

Given an array 'arr' and it's size 'n'
1) First traversal array element, i goes from 0 to n-1
a) Multiple each element with its index arr[i] = arr[i] * i
2) After that step 1. whole process is similar to Count Inversions in an array.

Below the implementation of above idea

C++

 // C++ program to count all pair that // hold condition i*arr[i] > j*arr[j] #include using namespace std;    /* This function merges two sorted arrays and    returns inversion count in the arrays.*/ int merge(int arr[], int temp[], int left,                        int mid, int right) {     int inv_count = 0;        int i = left; /* index for left subarray*/     int j = mid;  /* index for right subarray*/     int k = left; /* ndex for resultant subarray*/     while ((i <= mid - 1) && (j <= right))     {         if (arr[i] <= arr[j])             temp[k++] = arr[i++];         else         {             temp[k++] = arr[j++];                inv_count = inv_count + (mid - i);         }     }        /* Copy the remaining elements of left      subarray (if there are any) to temp*/     while (i <= mid - 1)         temp[k++] = arr[i++];        /* Copy the remaining elements of right      subarray (if there are any) to temp*/     while (j <= right)         temp[k++] = arr[j++];        /* Copy back the merged elements to original       array*/     for (i=left; i <= right; i++)         arr[i] = temp[i];        return inv_count; }    /* An auxiliary recursive function that sorts    the input array and returns the number of    inversions in the array. */ int _mergeSort(int arr[], int temp[], int left,                                       int right) {     int mid, inv_count = 0;     if (right > left)     {         /* Divide the array into two parts and call           _mergeSortAndCountInv() for each of           the parts */         mid = (right + left)/2;            /* Inversion count will be sum of inversions in            left-part, right-part and number of inversions            in merging */         inv_count  = _mergeSort(arr, temp, left, mid);         inv_count += _mergeSort(arr, temp, mid+1, right);            /*Merge the two parts*/         inv_count += merge(arr, temp, left, mid+1, right);     }        return inv_count; }    /* This function sorts the input array and    returns the number of inversions in the    array */ int countPairs(int arr[], int n) {     // Modify the array so that problem reduces to     // count inversion problem.     for (int i=0; i

Java

 // Java program to count all pair that // hold condition i*arr[i] > j*arr[j] import java.io.*;    class GFG  {     // This function merges two sorted arrays and     // returns inversion count in the arrays.     static int merge(int arr[], int temp[], int left,                                    int mid, int right)     {         int inv_count = 0;                    /* index for left subarray*/          int i = left;                     /* index for right subarray*/         int j = mid;          /* ndex for resultant subarray*/         int k = left;                     while ((i <= mid - 1) && (j <= right))         {             if (arr[i] <= arr[j])                 temp[k++] = arr[i++];             else             {                 temp[k++] = arr[j++];                        inv_count = inv_count + (mid - i);             }         }                /* Copy the remaining elements of left         subarray (if there are any) to temp*/         while (i <= mid - 1)             temp[k++] = arr[i++];                /* Copy the remaining elements of right         subarray (if there are any) to temp*/         while (j <= right)             temp[k++] = arr[j++];                // Copy back the merged elements          // to original array         for (i = left; i <= right; i++)             arr[i] = temp[i];                return inv_count;     }            /* An auxiliary recursive function      that sorts the input array and      returns the number of inversions      in the array. */     static int _mergeSort(int arr[], int temp[],                                 int left,int right)     {         int mid, inv_count = 0;         if (right > left)         {             /* Divide the array into two parts and call             _mergeSortAndCountInv() for each of             the parts */             mid = (right + left) / 2;                    // Inversion count will be sum of inversions in             // left-part, right-part and number of inversions             // in merging              inv_count = _mergeSort(arr, temp, left, mid);             inv_count += _mergeSort(arr, temp, mid+1, right);                    /*Merge the two parts*/             inv_count += merge(arr, temp, left, mid+1, right);         }                return inv_count;     }            // This function sorts the input array and     // returns the number of inversions in the     // array      static int countPairs(int arr[], int n)     {         // Modify the array so that problem reduces to         // count inversion problem.         for (int i = 0; i < n; i++)             arr[i] = i * arr[i];                // Count inversions using same logic as         // below post         int temp[] = new int [n];         return _mergeSort(arr, temp, 0, n - 1);     }            // Driver code        public static void main (String[] args)      {         int arr[] = {5, 0, 10, 2, 4, 1, 6};         int n = arr.length;         System.out.print( "Count of Pairs : "                           + countPairs(arr, n));                  } }    // This code is contributed by vt_m

Python 3

 # Python 3 program to count all  # pair that hold condition # i*arr[i] > j*arr[j]    # This function merges two  # sorted arrays and returns  # inversion count in the arrays. def merge(arr, temp, left, mid, right):        inv_count = 0        i = left # index for left subarray     j = mid # index for right subarray     k = left # ndex for resultant subarray     while ((i <= mid - 1) and (j <= right)):                if (arr[i] <= arr[j]):             temp[k] = arr[i]             i += 1             k += 1         else:                        temp[k] = arr[j]             k += 1             j += 1                inv_count = inv_count + (mid - i)        # Copy the remaining elements of left     # subarray (if there are any) to temp     while (i <= mid - 1):         temp[k] = arr[i]         i += 1         k += 1        # Copy the remaining elements of right     # subarray (if there are any) to temp     while (j <= right):         temp[k] = arr[j]         k += 1         j += 1        # Copy back the merged elements      # to original array     for i in range(left, right + 1):         arr[i] = temp[i]        return inv_count    # An auxiliary recursive function  # that sorts the input array and  # returns the number of inversions # in the array.  def _mergeSort(arr, temp, left, right):        inv_count = 0     if (right > left):                # Divide the array into two parts          # and call _mergeSortAndCountInv()          # for each of the parts          mid = (right + left) // 2            # Inversion count will be sum of          # inversions in left-part, right-part x         # and number of inversions in merging          inv_count = _mergeSort(arr, temp, left, mid)         inv_count += _mergeSort(arr, temp,                                  mid + 1, right)            # Merge the two parts         inv_count += merge(arr, temp, left,                                 mid + 1, right)        return inv_count    # This function sorts the input  # array and returns the number  # of inversions in the array  def countPairs(arr, n):            # Modify the array so that problem      # reduces to count inversion problem.     for i in range(n):         arr[i] = i * arr[i]        # Count inversions using same      # logic as below post     temp =  * n     return _mergeSort(arr, temp, 0, n - 1)    # Driver code if __name__ == "__main__":     arr = [5, 0, 10, 2, 4, 1, 6]     n = len(arr)     print("Count of Pairs : ",             countPairs(arr, n))               # This code is contributed  # by ChitraNayal

C#

 // C# program to count all pair that // hold condition i*arr[i] > j*arr[j] using System;    class GFG  {     // This function merges two sorted arrays and     // returns inversion count in the arrays.     static int merge(int []arr, int []temp, int left,                                 int mid, int right)     {         int inv_count = 0;                    /* index for left subarray*/         int i = left;                     /* index for right subarray*/         int j = mid;          /* ndex for resultant subarray*/         int k = left;                     while ((i <= mid - 1) && (j <= right))         {             if (arr[i] <= arr[j])                 temp[k++] = arr[i++];             else             {                 temp[k++] = arr[j++];                        inv_count = inv_count + (mid - i);             }         }                /* Copy the remaining elements of left         subarray (if there are any) to temp*/         while (i <= mid - 1)             temp[k++] = arr[i++];                /* Copy the remaining elements of right         subarray (if there are any) to temp*/         while (j <= right)             temp[k++] = arr[j++];                // Copy back the merged elements          // to original array         for (i = left; i <= right; i++)             arr[i] = temp[i];                return inv_count;     }            /* An auxiliary recursive function      that sorts the input array and      returns the number of inversions      in the array. */     static int _mergeSort(int []arr, int []temp,                              int left,int right)     {         int mid, inv_count = 0;         if (right > left)         {             /* Divide the array into two parts and call             _mergeSortAndCountInv() for each of             the parts */             mid = (right + left) / 2;                    // Inversion count will be sum of inversions in             // left-part, right-part and number of inversions             // in merging              inv_count = _mergeSort(arr, temp, left, mid);             inv_count += _mergeSort(arr, temp, mid+1, right);                    /*Merge the two parts*/             inv_count += merge(arr, temp, left, mid+1, right);         }                return inv_count;     }            // This function sorts the input array and     // returns the number of inversions in the     // array      static int countPairs(int []arr, int n)     {         // Modify the array so that problem reduces to         // count inversion problem.         for (int i = 0; i < n; i++)             arr[i] = i * arr[i];                // Count inversions using same logic as         // below post         int []temp = new int [n];         return _mergeSort(arr, temp, 0, n - 1);     }            // Driver code        public static void Main ()      {         int []arr = {5, 0, 10, 2, 4, 1, 6};         int n = arr.Length;         Console.WriteLine( "Count of Pairs : "                         + countPairs(arr, n));                 } }    // This code is contributed by anuj_67.

Output:

Count of Pairs : 5

Time Complexity: O(n log n)

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Improved By : Sam007, jit_t, vt_m, chitranayal