Count subsequences for every array element in which they are the maximum

Given an array arr[] consisting of N unique elements, the task is to generate an array B[] of length N such that B[i] is the number of subsequences in which arr[i] is the maximum element.

Examples:

Input: arr[] = {2, 3, 1}
Output: {2, 4, 1}
Explanation: Subsequences in which arr[0] ( = 2) is maximum are {2}, {2, 1}.
Subsequences in which arr[1] ( = 3) is maximum are {3}, {1, 3, 2}, {2, 3}, {1, 3}.
Subsequence in which arr[2] ( = 1) is maximum is {1}.

Input: arr[] = {23, 34, 12, 7, 15, 31}
Output: {8, 32, 2, 1, 4, 16}

Approach: The problem can be solved by observing that all the subsequences where an element, arr[i], will appear as the maximum will contain all the elements less than arr[i]. Therefore, the total number of distinct subsequences will be 2^{(Number of elements less than arr[i])}. Follow the steps below to solve the problem:

1. Sort the array arr[] indices with respect to their corresponding values present in the given array and store that indices in array indices[], where arr[indices[i]] < arr[indices[i+1]].
2. Initialize an integer, subseq with 1 to store the number of possible subsequences.
3. Iterate N times with pointer over the range [0, N-1] using a variable, i.
   1. B[indices[i]] is the number of subsequences in which arr[indices[i]] is maximum i.e., 2ⁱ, as there will be i elements less than arr[indices[i]].
   2. Store the answer for B[indices[i]] as B[indices[i]] = subseq.
   3. Update subseq by multiplying it by 2.
4. Print the elements of the array B[] as the answer.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to merge the subarrays
// arr[l .. m] and arr[m + 1, .. r]
// based on indices[]
void merge(int* indices, int* a, int l,
           int mid, int r)
{
    int temp_ind[r - l + 1], j = mid + 1;
    int i = 0, temp_l = l, k;
    while (l <= mid && j <= r) {

        // If a[indices[l]] is less than
        // a[indices[j]], add indice[l] to temp
        if (a[indices[l]] < a[indices[j]])
            temp_ind[i++] = indices[l++];

        // Else add indices[j]
        else
            temp_ind[i++] = indices[j++];
    }

    // Add remaining elements
    while (l <= mid)
        temp_ind[i++] = indices[l++];

    // Add remaining elements
    while (j <= r)
        temp_ind[i++] = indices[j++];
    for (k = 0; k < i; k++)
        indices[temp_l++] = temp_ind[k];
}

// Recursive function to divide
// the array into parts
void divide(int* indices, int* a, int l, int r)
{
    if (l >= r)
        return;
    int mid = l / 2 + r / 2;

    // Recursive call for elements before mid
    divide(indices, a, l, mid);

    // Recursive call for elements after mid
    divide(indices, a, mid + 1, r);

    // Merge the two sorted arrays
    merge(indices, a, l, mid, r);
}

// Function to find the number of
// subsequences for each element
void noOfSubsequences(int arr[], int N)
{
    int indices[N], i;
    for (i = 0; i < N; i++)
        indices[i] = i;

    // Sorting the indices according
    // to array arr[]
    divide(indices, arr, 0, N - 1);

    // Array to store output numbers
    int B[N];

    // Initialize subseq
    int subseq = 1;
    for (i = 0; i < N; i++) {

        // B[i] is 2^i
        B[indices[i]] = subseq;

        // Doubling the subsequences
        subseq *= 2;
    }
    // Print the final output, array B[]
    for (i = 0; i < N; i++)
        cout << B[i] << " ";
}

// Driver Code
int main()
{

    // Given array
    int arr[] = { 2, 3, 1 };

    // Given length
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    noOfSubsequences(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to merge the subarrays
// arr[l .. m] and arr[m + 1, .. r]
// based on indices[]
static void merge(int[] indices, int[] a, int l,
                  int mid, int r)
{
    int []temp_ind = new int[r - l + 1];
    int j = mid + 1;
    int i = 0, temp_l = l, k;

    while (l <= mid && j <= r)
    {

        // If a[indices[l]] is less than
        // a[indices[j]], add indice[l] to temp
        if (a[indices[l]] < a[indices[j]])
            temp_ind[i++] = indices[l++];

        // Else add indices[j]
        else
            temp_ind[i++] = indices[j++];
    }

    // Add remaining elements
    while (l <= mid)
        temp_ind[i++] = indices[l++];

    // Add remaining elements
    while (j <= r)
        temp_ind[i++] = indices[j++];

    for(k = 0; k < i; k++)
        indices[temp_l++] = temp_ind[k];
}

// Recursive function to divide
// the array into parts
static void divide(int[] indices, int[] a,
                   int l, int r)
{
    if (l >= r)
        return;

    int mid = l / 2 + r / 2;

    // Recursive call for elements before mid
    divide(indices, a, l, mid);

    // Recursive call for elements after mid
    divide(indices, a, mid + 1, r);

    // Merge the two sorted arrays
    merge(indices, a, l, mid, r);
}

// Function to find the number of
// subsequences for each element
static void noOfSubsequences(int arr[], int N)
{
    int []indices = new int[N];
    int i;

    for(i = 0; i < N; i++)
        indices[i] = i;

    // Sorting the indices according
    // to array arr[]
    divide(indices, arr, 0, N - 1);

    // Array to store output numbers
    int[] B = new int[N];

    // Initialize subseq
    int subseq = 1;

    for(i = 0; i < N; i++)
    {

        // B[i] is 2^i
        B[indices[i]] = subseq;

        // Doubling the subsequences
        subseq *= 2;
    }

    // Print the final output, array B[]
    for(i = 0; i < N; i++)
        System.out.print(B[i] + " ");
}

// Driver Code
public static void main(String[] args)
{

    // Given array
    int arr[] = { 2, 3, 1 };

    // Given length
    int N = arr.length;

    // Function call
    noOfSubsequences(arr, N);
}
}

// This code is contributed by Princi Singh

Python3

# Python3 program for the above approach

# Function to merge the subarrays
# arr[l .. m] and arr[m + 1, .. r]
# based on indices[]
def merge(indices, a, l, mid, r):

    temp_ind = [0] * (r - l + 1)
    j = mid + 1
    i = 0
    temp_l = l

    while (l <= mid and j <= r):

        # If a[indices[l]] is less than
        # a[indices[j]], add indice[l] to temp
        if (a[indices[l]] < a[indices[j]]):
            temp_ind[i] = indices[l]
            i += 1
            l += 1

        # Else add indices[j]
        else:
            temp_ind[i] = indices[j]
            i += 1
            j += 1

    # Add remaining elements
    while (l <= mid):
        temp_ind[i] = indices[l]
        i += 1
        l += 1

    # Add remaining elements
    while (j <= r):
        temp_ind[i] = indices[j]
        i += 1
        j += 1

    for k in range(i):
        indices[temp_l] = temp_ind[k]
        temp_l += 1

# Recursive function to divide
# the array into parts
def divide(indices, a, l, r):

    if (l >= r):
        return

    mid = l // 2 + r // 2

    # Recursive call for elements
    # before mid
    divide(indices, a, l, mid)

    # Recursive call for elements
    # after mid
    divide(indices, a, mid + 1, r)

    # Merge the two sorted arrays
    merge(indices, a, l, mid, r)

# Function to find the number of
# subsequences for each element
def noOfSubsequences(arr, N):

    indices = N * [0]
    for i in range(N):
        indices[i] = i

    # Sorting the indices according
    # to array arr[]
    divide(indices, arr, 0, N - 1)

    # Array to store output numbers
    B = [0] * N

    # Initialize subseq
    subseq = 1
    for i in range(N):

        # B[i] is 2^i
        B[indices[i]] = subseq

        # Doubling the subsequences
        subseq *= 2

    # Print the final output, array B[]
    for i in range(N):
        print(B[i], end = " ")

# Driver Code
if __name__ == "__main__":

    # Given array
    arr = [ 2, 3, 1 ]

    # Given length
    N = len(arr)

    # Function call
    noOfSubsequences(arr, N)

# This code is contributed by chitranayal

C#

// C# program for the above approach
using System;

class GFG{

// Function to merge the subarrays
// arr[l .. m] and arr[m + 1, .. r]
// based on indices[]
static void merge(int[] indices, int[] a, int l,
                  int mid, int r)
{
    int []temp_ind = new int[r - l + 1];
    int j = mid + 1;
    int i = 0, temp_l = l, k;

    while (l <= mid && j <= r)
    {

        // If a[indices[l]] is less than
        // a[indices[j]], add indice[l] to temp
        if (a[indices[l]] < a[indices[j]])
            temp_ind[i++] = indices[l++];

        // Else add indices[j]
        else
            temp_ind[i++] = indices[j++];
    }

    // Add remaining elements
    while (l <= mid)
        temp_ind[i++] = indices[l++];

    // Add remaining elements
    while (j <= r)
        temp_ind[i++] = indices[j++];

    for(k = 0; k < i; k++)
        indices[temp_l++] = temp_ind[k];
}

// Recursive function to divide
// the array into parts
static void divide(int[] indices, int[] a,
                   int l, int r)
{
    if (l >= r)
        return;

    int mid = l / 2 + r / 2;

    // Recursive call for elements before mid
    divide(indices, a, l, mid);

    // Recursive call for elements after mid
    divide(indices, a, mid + 1, r);

    // Merge the two sorted arrays
    merge(indices, a, l, mid, r);
}

// Function to find the number of
// subsequences for each element
static void noOfSubsequences(int []arr, int N)
{
    int []indices = new int[N];
    int i;

    for(i = 0; i < N; i++)
        indices[i] = i;

    // Sorting the indices according
    // to array []arr
    divide(indices, arr, 0, N - 1);

    // Array to store output numbers
    int[] B = new int[N];

    // Initialize subseq
    int subseq = 1;

    for(i = 0; i < N; i++)
    {

        // B[i] is 2^i
        B[indices[i]] = subseq;

        // Doubling the subsequences
        subseq *= 2;
    }

    // Print the readonly output, array []B
    for(i = 0; i < N; i++)
        Console.Write(B[i] + " ");
}

// Driver Code
public static void Main(String[] args)
{

    // Given array
    int []arr = { 2, 3, 1 };

    // Given length
    int N = arr.Length;

    // Function call
    noOfSubsequences(arr, N);
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// JavaScript program for the above approach

// Function to merge the subarrays
// arr[l .. m] and arr[m + 1, .. r]
// based on indices[]
function merge(indices, a, l, mid, r)
{
    let temp_ind = [];
    let j = mid + 1;
    let i = 0, temp_l = l, k;

    while (l <= mid && j <= r)
    {

        // If a[indices[l]] is less than
        // a[indices[j]], add indice[l] to temp
        if (a[indices[l]] < a[indices[j]])
            temp_ind[i++] = indices[l++];

        // Else add indices[j]
        else
            temp_ind[i++] = indices[j++];
    }

    // Add remaining elements
    while (l <= mid)
        temp_ind[i++] = indices[l++];

    // Add remaining elements
    while (j <= r)
        temp_ind[i++] = indices[j++];

    for(k = 0; k < i; k++)
        indices[temp_l++] = temp_ind[k];
}

// Recursive function to divide
// the array into parts
function divide(indices, a, l, r)
{
    if (l >= r)
        return;

    let mid = l / 2 + r / 2;

    // Recursive call for elements before mid
    divide(indices, a, l, mid);

    // Recursive call for elements after mid
    divide(indices, a, mid + 1, r);

    // Merge the two sorted arrays
    merge(indices, a, l, mid, r);
}

// Function to find the number of
// subsequences for each element
function noOfSubsequences(arr, N)
{
    let indices = [];
    let i;

    for(i = 0; i < N; i++)
        indices[i] = i;

    // Sorting the indices according
    // to array arr[]
    divide(indices, arr, 0, N - 1);

    // Array to store output numbers
    let B = [];

    // Initialize subseq
    let subseq = 1;

    for(i = 0; i < N; i++)
    {

        // B[i] is 2^i
        B[indices[i]] = subseq;

        // Doubling the subsequences
        subseq *= 2;
    }

    // Print the final output, array B[]
    for(i = 0; i < N; i++)
        document.write(B[i] + " ");
}

// Driver code

// Given array
let arr = [ 2, 3, 1 ];

// Given length
let N = arr.length;

// Function call
noOfSubsequences(arr, N);

// This code is contributed by avijitmondal1998

</script>

Output:

2 4 1

Time Complexity: O(NlogN) where N is the length of the given array.
Auxiliary Space: O(N)