Count Inversions of size three in a given array
Last Updated :
20 Mar, 2023
Given an array arr[] of size n. Three elements arr[i], arr[j] and arr[k] form an inversion of size 3 if a[i] > a[j] >a[k] and i < j < k. Find total number of inversions of size 3.
Example :
Input: {8, 4, 2, 1}
Output: 4
The four inversions are (8,4,2), (8,4,1), (4,2,1) and (8,2,1).
Input: {9, 6, 4, 5, 8}
Output: 2
The two inversions are {9, 6, 4} and {9, 6, 5}
We have already discussed inversion count of size two by merge sort, Self Balancing BST and BIT.
Simple approach: Loop for all possible value of i, j and k and check for the condition a[i] > a[j] > a[k] and i < j < k.
C++
// A Simple C++ O(n^3) program to count inversions of size 3
#include<bits/stdc++.h>
using namespace std;
// Returns counts of inversions of size three
int getInvCount(int arr[],int n)
{
int invcount = 0; // Initialize result
for (int i=0; i<n-2; i++)
{
for (int j=i+1; j<n-1; j++)
{
if (arr[i]>arr[j])
{
for (int k=j+1; k<n; k++)
{
if (arr[j]>arr[k])
invcount++;
}
}
}
}
return invcount;
}
// Driver program to test above function
int main()
{
int arr[] = {8, 4, 2, 1};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Inversion Count : " << getInvCount(arr, n);
return 0;
}
// A simple Java implementation to count inversion of size 3
class Inversion{
// returns count of inversion of size 3
int getInvCount(int arr[], int n)
{
int invcount = 0; // initialize result
for(int i=0 ; i< n-2; i++)
{
for(int j=i+1; j<n-1; j++)
{
if(arr[i] > arr[j])
{
for(int k=j+1; k<n; k++)
{
if(arr[j] > arr[k])
invcount++;
}
}
}
}
return invcount;
}
// driver program to test above function
public static void main(String args[])
{
Inversion inversion = new Inversion();
int arr[] = new int[] {8, 4, 2, 1};
int n = arr.length;
System.out.print("Inversion count : " +
inversion.getInvCount(arr, n));
}
}
// This code is contributed by Mayank Jaiswal
# A simple python O(n^3) program
# to count inversions of size 3
# Returns counts of inversions
# of size threee
def getInvCount(arr):
n = len(arr)
invcount = 0 #Initialize result
for i in range(0,n-1):
for j in range(i+1 , n):
if arr[i] > arr[j]:
for k in range(j+1 , n):
if arr[j] > arr[k]:
invcount += 1
return invcount
# Driver program to test above function
arr = [8 , 4, 2 , 1]
print ("Inversion Count : %d" %(getInvCount(arr)))
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
// A simple C# implementation to
// count inversion of size 3
using System;
class GFG {
// returns count of inversion of size 3
static int getInvCount(int []arr, int n)
{
// initialize result
int invcount = 0;
for(int i = 0 ; i < n - 2; i++)
{
for(int j = i + 1; j < n - 1; j++)
{
if(arr[i] > arr[j])
{
for(int k = j + 1; k < n; k++)
{
if(arr[j] > arr[k])
invcount++;
}
}
}
}
return invcount;
}
// Driver Code
public static void Main()
{
int []arr = new int[] {8, 4, 2, 1};
int n = arr.Length;
Console.WriteLine("Inversion count : " +
getInvCount(arr, n));
}
}
// This code is contributed by anuj_67.
<?php
// A O(n^2) PHP program to
// count inversions of size 3
// Returns count of
// inversions of size 3
function getInvCount($arr, $n)
{
// Initialize result
$invcount = 0;
for ($i = 1; $i < $n - 1; $i++)
{
// Count all smaller elements
// on right of arr[i]
$small = 0;
for($j = $i + 1; $j < $n; $j++)
if ($arr[$i] > $arr[$j])
$small++;
// Count all greater elements
// on left of arr[i]
$great = 0;
for($j = $i - 1; $j >= 0; $j--)
if ($arr[$i] < $arr[$j])
$great++;
// Update inversion count by
// adding all inversions
// that have arr[i] as
// middle of three elements
$invcount += $great * $small;
}
return $invcount;
}
// Driver Code
$arr = array(8, 4, 2, 1);
$n = sizeof($arr);
echo "Inversion Count : "
, getInvCount($arr, $n);
// This code is contributed m_kit
?>
<script>
// A simple Javascript implementation to count inversion of size 3
// returns count of inversion of size 3
function getInvCount(arr, n)
{
let invcount = 0; // initialize result
for(let i = 0 ; i < n - 2; i++)
{
for(let j = i + 1; j < n - 1; j++)
{
if(arr[i] > arr[j])
{
for(let k = j + 1; k < n; k++)
{
if(arr[j] > arr[k])
invcount++;
}
}
}
}
return invcount;
}
// driver program to test above function
let arr = [8, 4, 2, 1];
let n = arr.length;
document.write("Inversion count : " +
getInvCount(arr, n));
// This code is contributed by rag2127
</script>
OutputInversion Count : 4
Time complexity: O(n^3)
Auxiliary Space: O(1).
Better Approach :
We can reduce the complexity if we consider every element arr[i] as middle element of inversion, find all the numbers greater than a[i] whose index is less than i, find all the numbers which are smaller than a[i] and index is more than i. We multiply the number of elements greater than a[i] to the number of elements smaller than a[i] and add it to the result.
Below is the implementation of the idea.
C++
// A O(n^2) C++ program to count inversions of size 3
#include<bits/stdc++.h>
using namespace std;
// Returns count of inversions of size 3
int getInvCount(int arr[], int n)
{
int invcount = 0; // Initialize result
for (int i=1; i<n-1; i++)
{
// Count all smaller elements on right of arr[i]
int small = 0;
for (int j=i+1; j<n; j++)
if (arr[i] > arr[j])
small++;
// Count all greater elements on left of arr[i]
int great = 0;
for (int j=i-1; j>=0; j--)
if (arr[i] < arr[j])
great++;
// Update inversion count by adding all inversions
// that have arr[i] as middle of three elements
invcount += great*small;
}
return invcount;
}
// Driver program to test above function
int main()
{
int arr[] = {8, 4, 2, 1};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Inversion Count : " << getInvCount(arr, n);
return 0;
}
// A O(n^2) Java program to count inversions of size 3
class Inversion {
// returns count of inversion of size 3
int getInvCount(int arr[], int n)
{
int invcount = 0; // initialize result
for (int i=0 ; i< n-1; i++)
{
// count all smaller elements on right of arr[i]
int small=0;
for (int j=i+1; j<n; j++)
if (arr[i] > arr[j])
small++;
// count all greater elements on left of arr[i]
int great = 0;
for (int j=i-1; j>=0; j--)
if (arr[i] < arr[j])
great++;
// update inversion count by adding all inversions
// that have arr[i] as middle of three elements
invcount += great*small;
}
return invcount;
}
// driver program to test above function
public static void main(String args[])
{
Inversion inversion = new Inversion();
int arr[] = new int[] {8, 4, 2, 1};
int n = arr.length;
System.out.print("Inversion count : " +
inversion.getInvCount(arr, n));
}
}
// This code has been contributed by Mayank Jaiswal
# A O(n^2) Python3 program to
# count inversions of size 3
# Returns count of inversions
# of size 3
def getInvCount(arr, n):
# Initialize result
invcount = 0
for i in range(1,n-1):
# Count all smaller elements
# on right of arr[i]
small = 0
for j in range(i+1 ,n):
if (arr[i] > arr[j]):
small+=1
# Count all greater elements
# on left of arr[i]
great = 0;
for j in range(i-1,-1,-1):
if (arr[i] < arr[j]):
great+=1
# Update inversion count by
# adding all inversions that
# have arr[i] as middle of
# three elements
invcount += great * small
return invcount
# Driver program to test above function
arr = [8, 4, 2, 1]
n = len(arr)
print("Inversion Count :",getInvCount(arr, n))
# This code is Contributed by Smitha Dinesh Semwal
// A O(n^2) Java program to count inversions
// of size 3
using System;
public class Inversion {
// returns count of inversion of size 3
static int getInvCount(int []arr, int n)
{
int invcount = 0; // initialize result
for (int i = 0 ; i < n-1; i++)
{
// count all smaller elements on
// right of arr[i]
int small = 0;
for (int j = i+1; j < n; j++)
if (arr[i] > arr[j])
small++;
// count all greater elements on
// left of arr[i]
int great = 0;
for (int j = i-1; j >= 0; j--)
if (arr[i] < arr[j])
great++;
// update inversion count by
// adding all inversions that
// have arr[i] as middle of
// three elements
invcount += great * small;
}
return invcount;
}
// driver program to test above function
public static void Main()
{
int []arr = new int[] {8, 4, 2, 1};
int n = arr.Length;
Console.WriteLine("Inversion count : "
+ getInvCount(arr, n));
}
}
// This code has been contributed by anuj_67.
<?php
// A O(n^2) PHP program to count
// inversions of size 3
// Returns count of
// inversions of size 3
function getInvCount($arr, $n)
{
// Initialize result
$invcount = 0;
for ($i = 1; $i < $n - 1; $i++)
{
// Count all smaller elements
// on right of arr[i]
$small = 0;
for ($j = $i + 1; $j < $n; $j++)
if ($arr[$i] > $arr[$j])
$small++;
// Count all greater elements
// on left of arr[i]
$great = 0;
for ($j = $i - 1; $j >= 0; $j--)
if ($arr[$i] < $arr[$j])
$great++;
// Update inversion count by
// adding all inversions that
// have arr[i] as middle of
// three elements
$invcount += $great * $small;
}
return $invcount;
}
// Driver Code
$arr = array (8, 4, 2, 1);
$n = sizeof($arr);
echo "Inversion Count : " ,
getInvCount($arr, $n);
// This code is contributed by m_kit
?>
<script>
// A O(n^2) Javascript program to count inversions of size 3
// returns count of inversion of size 3
function getInvCount(arr, n)
{
let invcount = 0; // initialize result
for (let i = 0 ; i < n - 1; i++)
{
// count all smaller elements on right of arr[i]
let small = 0;
for (let j = i + 1; j < n; j++)
if (arr[i] > arr[j])
small++;
// count all greater elements on left of arr[i]
let great = 0;
for (let j = i - 1; j >= 0; j--)
if (arr[i] < arr[j])
great++;
// update inversion count by adding all inversions
// that have arr[i] as middle of three elements
invcount += great*small;
}
return invcount;
}
// driver program to test above function
let arr=[8, 4, 2, 1];
let n = arr.length;
document.write("Inversion count : " +getInvCount(arr, n));
// This code is contributed by avanitrachhadiya2155
</script>
OutputInversion Count : 4
Time Complexity: O(n^2)
Auxiliary Space: O(1).
Binary Indexed Tree Approach :
Like inversions of size 2, we can use Binary indexed tree to find inversions of size 3. It is strongly recommended to refer below article first.
Count inversions of size two Using BIT
The idea is similar to above method. We count the number of greater elements and smaller elements for all the elements and then multiply greater[] to smaller[] and add it to the result.
Solution :
- To find out the number of smaller elements for an index we iterate from n-1 to 0. For every element a[i] we calculate the getSum() function for (a[i]-1) which gives the number of elements till a[i]-1.
- To find out the number of greater elements for an index we iterate from 0 to n-1. For every element a[i] we calculate the sum of numbers till a[i] (sum smaller or equal to a[i]) by getSum() and subtract it from i (as i is the total number of element till that point) so that we can get number of elements greater than a[i].
Below is the code for the above approach.
C++
// C++ program to count inversions of size 3 using
// Binary Indexed Tree
#include <bits/stdc++.h>
using namespace std;
// It is beneficial to declare the 2D BIT globally
// since passing it into functions will create
// additional overhead
const int N = 100005;
int BIT[3][N] = { 0 };
// update function. "t" denotes the t'th Binary
// indexed tree
void updateBIT(int t, int i, int val, int n)
{
// Traversing the t'th BIT
while (i <= n) {
BIT[t][i] = BIT[t][i] + val;
i = i + (i & (-i));
}
}
// function to get the sum.
// "t" denotes the t'th Binary indexed tree
int getSum(int t, int i)
{
int res = 0;
// Traversing the t'th BIT
while (i > 0) {
res = res + BIT[t][i];
i = i - (i & (-i));
}
return res;
}
// Converts an array to an array with values from 1 to n
// and relative order of smaller and greater elements
// remains same.
void convert(int arr[], int n)
{
// Create a copy of arr[] in temp and sort
// the temp array in increasing order
int temp[n];
for (int i = 0; i < n; i++)
temp[i] = arr[i];
sort(temp, temp + n);
// Traverse all array elements
for (int i = 0; i < n; i++) {
// lower_bound() Returns pointer to the
// first element greater than or equal
// to arr[i]
arr[i] = lower_bound(temp, temp + n,
arr[i]) - temp + 1;
}
}
// Returns count of inversions of size three
int getInvCount(int arr[], int n)
{
// Convert arr[] to an array with values from
// 1 to n and relative order of smaller and
// greater elements remains same.
convert(arr, n);
// iterating over the converted array in
// reverse order.
for (int i = n - 1; i >= 0; i--) {
// update the BIT for l = 1
updateBIT(1, arr[i], 1, n);
// update BIT for all other BITs
for (int l = 1; l < 3; l++) {
updateBIT(l + 1, arr[i], getSum(l, arr[i] - 1), n);
}
}
// final result
return getSum(3, n);
}
// Driver program to test above function
int main()
{
int arr[] = {8, 4, 2, 1};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Inversion Count : " << getInvCount(arr, n);
return 0;
}
// This code is contributed by Pushpesh Raj.
// Java program to count inversions of size 3 using
// Binary Indexed Tree
import java.io.*;
import java.util.Arrays;
import java.util.ArrayList;
import java.lang.*;
import java.util.Collections;
class GFG {
// It is beneficial to declare the 2D BIT globally
// since passing it into functions will create
// additional overhead
static int N = 100005;
static int BIT[][] = new int[4][N];
// update function. "t" denotes the t'th Binary
// indexed tree
static void updateBIT(int t, int i, int val, int n)
{
// Traversing the t'th BIT
while (i <= n) {
BIT[t][i] = BIT[t][i] + val;
i = i + (i & (-i));
}
}
// function to get the sum.
// "t" denotes the t'th Binary indexed tree
static int getSum(int t, int i)
{
int res = 0;
// Traversing the t'th BIT
while (i > 0) {
res = res + BIT[t][i];
i = i - (i & (-i));
}
return res;
}
// Converts an array to an array with values from 1 to n
// and relative order of smaller and greater elements
// remains same.
static void convert(int arr[], int n)
{
// Create a copy of arr[] in temp and sort
// the temp array in increasing order
int temp[]=new int[n];
for (int i = 0; i < n; i++)
temp[i] = arr[i];
Arrays.sort(temp);
// Traverse all array elements
for (int i = 0; i < n; i++) {
// lower_bound() Returns pointer to the
// first element greater than or equal
// to arr[i]
arr[i] = Arrays.binarySearch(temp,arr[i]) + 1;
}
}
// Returns count of inversions of size three
static int getInvCount(int arr[], int n)
{
// Convert arr[] to an array with values from
// 1 to n and relative order of smaller and
// greater elements remains same.
convert(arr, n);
// iterating over the converted array in
// reverse order.
for (int i = n - 1; i >= 0; i--) {
// update the BIT for l = 1
updateBIT(1, arr[i], 1, n);
// update BIT for all other BITs
for (int l = 1; l < 3; l++) {
updateBIT(l + 1, arr[i], getSum(l, arr[i] - 1), n);
}
}
// final result
return getSum(3, n);
}
// Driver program to test above function
public static void main (String[] args)
{
int arr[] = {8, 4, 2, 1};
int n = arr.length;
System.out.print("Inversion Count : "+getInvCount(arr, n));
}
}
// This code is contributed by Utkarsh
# Python program to count inversions of size 3 using
# Binary Indexed Tree
# It is beneficial to declare the 2D BIT globally
# since passing it into functions will create
# additional overhead
N = 100005
BIT = [[0 for x in range(N)] for y in range(3)]
for i in range(0,50):
abc = [0]*50
BIT.append(abc)
# update function. "t" denotes the t'th Binary
# indexed tree
def updateBIT(t, i, val, n):
# Traversing the t'th BIT
while i <= n:
BIT[t][i] = BIT[t][i] + val
i = i + (i & (-i))
# function to get the sum.
# "t" denotes the t'th Binary indexed tree
def getSum(t, i):
res = 0
# Traversing the t'th BIT
while i > 0:
res = res + BIT[t][i]
i = i - (i & (-i))
return res
# Converts an array to an array with values from 1 to n
# and relative order of smaller and greater elements
# remains same.
def convert(arr, n):
# Create a copy of arr[] in temp and sort
# the temp array in increasing order
temp = [0 for x in range(n)]
for i in range(n):
temp[i] = arr[i]
temp.sort()
# Traverse all array elements
for i in range(n):
# lower_bound() Returns pointer to the
# first element greater than or equal
# to arr[i]
arr[i] = temp.index(arr[i]) + 1
# Returns count of inversions of size three
def getInvCount(arr, n):
# Convert arr[] to an array with values from
# 1 to n and relative order of smaller and
# greater elements remains same.
convert(arr, n)
# iterating over the converted array in
# reverse order.
for i in range(n - 1, -1, -1):
# update the BIT for l = 1
updateBIT(1, arr[i], 1, n)
# update BIT for all other BITs
for l in range(1, 3):
updateBIT(l + 1, arr[i], getSum(l, arr[i] - 1), n)
# final result
return getSum(3, n)
# Driver program to test above function
arr = [8, 4, 2, 1]
n = len(arr)
print("Inversion Count : ", getInvCount(arr, n))
# This code is contributed by ashishak__
// JavaScript program to count inversions of size 3 using
// Binary Indexed Tree
const N = 100005;
let BIT = [[0], [0], [0]];
for (let i = 0; i < 10; i++) {
let abc = [0];
BIT.push(abc);
}
// update function. "t" denotes the t'th Binary
// indexed tree
function updateBIT(t, i, val, n) {
// Traversing the t'th BIT
while (i <= n) {
if (!BIT[t][i]) BIT[t][i] = 0;
BIT[t][i] = BIT[t][i] + val;
i = i + (i & (-i));
}
}
// function to get the sum.
// "t" denotes the t'th Binary indexed tree
function getSum(t, i) {
let res = 0;
// Traversing the t'th BIT
while (i > 0) {
res = res + BIT[t][i];
i = i - (i & (-i));
}
return res;
}
// Converts an array to an array with values from 1 to n
// and relative order of smaller and greater elements
// remains same.
function convert(arr, n) {
// Create a copy of arr[] in temp and sort
// the temp array in increasing order
let temp = [];
for (let i = 0; i < n; i++)
temp[i] = arr[i];
temp.sort();
// Traverse all array elements
for (let i = 0; i < n; i++) {
// lower_bound() Returns pointer to the
// first element greater than or equal
// to arr[i]
arr[i] = temp.indexOf(arr[i]) + 1;
}
}
// Returns count of inversions of size three
function getInvCount(arr, n) {
// Convert arr[] to an array with values from
// 1 to n and relative order of smaller and
// greater elements remains same.
convert(arr, n);
// iterating over the converted array in
// reverse order.
for (let i = n - 1; i >= 0; i--) {
// update the BIT for l = 1
updateBIT(1, arr[i], 1, n);
// update BIT for all other BITs
for (let l = 1; l < 3; l++) {
updateBIT(l + 1, arr[i], getSum(l, arr[i] - 1), n);
}
}
// final result
return getSum(3, n);
}
// Driver program to test above function
let arr = [8, 4, 2, 1];
let n = arr.length;
console.log("Inversion Count : " + getInvCount(arr, n));
// This code is contributed by akashish__.
using System;
class GFG
{
// It is beneficial to declare the 2D BIT globally
// since passing it into functions will create
// additional overhead
static int N = 100005;
static int[,] BIT = new int[4, N];
// update function. "t" denotes the t'th Binary
// indexed tree
static void updateBIT(int t, int i, int val, int n)
{
// Traversing the t'th BIT
while (i <= n)
{
BIT[t, i] = BIT[t, i] + val;
i = i + (i & (-i));
}
}
// function to get the sum.
// "t" denotes the t'th Binary indexed tree
static int getSum(int t, int i)
{
int res = 0;
// Traversing the t'th BIT
while (i > 0)
{
res = res + BIT[t, i];
i = i - (i & (-i));
}
return res;
}
// Converts an array to an array with values from 1 to n
// and relative order of smaller and greater elements
// remains same.
static void convert(int[] arr, int n)
{
// Create a copy of arr[] in temp and sort
// the temp array in increasing order
int[] temp = new int[n];
for (int i = 0; i < n; i++)
temp[i] = arr[i];
Array.Sort(temp);
// Traverse all array elements
for (int i = 0; i < n; i++)
{
// lower_bound() Returns pointer to the
// first element greater than or equal
// to arr[i]
arr[i] = Array.BinarySearch(temp, arr[i]) + 1;
}
}
// Returns count of inversions of size three
static int getInvCount(int[] arr, int n)
{
// Convert arr[] to an array with values from
// 1 to n and relative order of smaller and
// greater elements remains same.
convert(arr, n);
// iterating over the converted array in
// reverse order.
for (int i = n - 1; i >= 0; i--)
{
// update the BIT for l = 1
updateBIT(1, arr[i], 1, n);
// update BIT for all other BITs
for (int l = 1; l < 3; l++)
{
updateBIT(l + 1, arr[i], getSum(l, arr[i] - 1), n);
}
}
// final result
return getSum(3, n);
}
// Driver program to test above function
static void Main(string[] args)
{
int[] arr = { 8, 4, 2, 1 };
int n = arr.Length;
Console.Write("Inversion Count : " + getInvCount(arr, n));
}
}
OutputInversion Count : 4
Time Complexity: O(n*log(n))
Auxiliary Space: O(n).
Similar Reads
Javascript Program to Count Inversions of size three in a given array
Given an array arr[] of size n. Three elements arr[i], arr[j] and arr[k] form an inversion of size 3 if a[i] > a[j] >a[k] and i < j < k. Find total number of inversions of size 3.Example : Input: {8, 4, 2, 1}Output: 4The four inversions are (8,4,2), (8,4,1), (4,2,1) and (8,2,1).Input: {9
4 min read
Count of Integers in given Array whose MSB and LSB are set
Given an array A[] of length N, the task is to count the number of elements of the array having their MSB and LSB set. Examples: Input: A[] = {2, 3, 1, 5}Output: 3Explanation: Binary representation of the array elements {2, 3, 1, 5} are {10, 11, 1, 101}The integers 3, 1, 5 are the integers whose f
9 min read
Count 1s present in a range of indices [L, R] in a given array
Given an array arr[] consisting of a single element N (1 ≤ N ≤ 106) and two integers L and R, ( 1 ≤ L ≤ R ≤ 105), the task is to make all array elements either 0 or 1 using the following operations : Select an element P such that P > 1 from the array arr[].Replace P with three elements at the sam
10 min read
Count of triplets in a given Array having GCD K
Given an integer array arr[] and an integer K, the task is to count all triplets whose GCD is equal to K.Examples: Input: arr[] = {1, 4, 8, 14, 20}, K = 2 Output: 3 Explanation: Triplets (4, 14, 20), (8, 14, 20) and (4, 8, 14) have GCD equal to 2.Input: arr[] = {1, 2, 3, 4, 5}, K = 7 Output: 0 Appro
8 min read
3 Sum - Count Triplets With Given Sum In Sorted Array
Given a sorted array arr[] and a target value, the task is to find the count of triplets present in the given array having sum equal to the given target. More specifically, the task is to count triplets (i, j, k) of valid indices, such that arr[i] + arr[j] + arr[k] = target and i < j < k.Examp
10 min read
Count subsequence of length three in a given string
Given a string of length n and a subsequence of length 3. Find the total number of occurrences of the subsequence in this string. Examples : Input : string = "GFGFGYSYIOIWIN", subsequence = "GFG" Output : 4 Explanation : There are 4 such subsequences as shown: GFGFGYSYIOIWIN GFGFGYSYIOIWIN GFGFGYSYI
15 min read
Count subarrays with non-zero sum in the given Array
Given an array arr[] of size N, the task is to count the total number of subarrays for the given array arr[] which have a non-zero-sum.Examples: Input: arr[] = {-2, 2, -3} Output: 4 Explanation: The subarrays with non zero sum are: [-2], [2], [2, -3], [-3]. All possible subarray of the given input a
7 min read
Count of triplets in an Array with odd sum
Given an array arr[] with N integers, find the number of triplets of i, j and k such that 1<= i < j < k <= N and arr[i] + arr[j] + arr[k] is odd. Example: Input: arr[] = {1, 2, 3, 4, 5}Output: 4Explanation: The given array contains 4 triplets with an odd sum. They are {1, 2, 4}, {1, 3, 5
6 min read
Count of subsequences having odd Bitwise AND values in the given array
Given an array arr[] of N integers, the task is to find the number of subsequences of the given array such that their Bitwise AND value is Odd. Examples: Input: arr[] = {2, 3, 1}Output: 3Explanation: The subsequences of the given array having odd Bitwise AND values are {3} = 3, {1} = 1, {3, 1} = 3
5 min read
Count of distinct alternate triplets of indices from given Array | Set 2
Given a binary array arr[] of size N, the task is to find the count of distinct alternating triplets. Note: A triplet is alternating if the values of those indices are in {0, 1, 0} or {1, 0, 1} form. Examples: Input: arr[] = {0, 0, 1, 1, 0, 1} Output: 6Explanation: Here four sequence of "010" and tw
9 min read