Given an arrays arr[] in which initially all elements are 0 and another array Q[][] containing K queries where every query represents a range [L, R], the task is to add 1 to each subarrays where each subarray is defined by the range [L, R], and return sum of all unique elements.
Note: One-based indexing is used in the Q[][] array to signify the ranges.
Examples:
Input: arr[] = { 0, 0, 0, 0, 0, 0 }, Q[][2] = {{1, 3}, {4, 6}, {3, 4}, {3, 3}}
Output: 6
Explanation:
Initially the array is arr[] = { 0, 0, 0, 0, 0, 0 }.
Query 1: arr[] = { 1, 1, 1, 0, 0, 0 }.
Query 2: arr[] = { 1, 1, 1, 1, 1, 1 }.
Query 3: arr[] = { 1, 1, 2, 2, 1, 1 }.
Query 4: arr[] = { 1, 1, 3, 2, 1, 1 }.
Hence unique elements are {1, 3, 2}. Thus sum = 1 + 3 + 2 = 6.
Input: arr[] = { 0, 0, 0, 0, 0, 0, 0, 0 }, Q[][2] = {{1, 4}, {5, 5}, {7, 8}, {8, 8}}
Output: 3
Explanation:
Initially the array is arr[] = { 0, 0, 0, 0, 0, 0, 0, 0 }.
After processing all queries, arr[] = { 1, 1, 1, 1, 1, 0, 1, 2 }.
Hence unique elements are {1, 0, 2}. Thus sum = 1 + 0 + 2 = 3.
Approach: The idea is to increment the value by 1 and decrement the array by 1 at L and R+1 index respectively for processing each query. Then, Compute the prefix sum of the array to find the final array after Q queries. As explained in this article. Finally, compute the sum of the unique elements with the help of the hash-map.
Below is the implementation of the above approach:
// C++ implementation to find the // sum of all unique elements of // the array after Q queries #include <bits/stdc++.h> using namespace std;
// Function to find the sum of // unique elements after Q Query int uniqueSum( int A[], int R[][2],
int N, int M)
{ // Updating the array after
// processing each query
for ( int i = 0; i < M; ++i) {
int l = R[i][0], r = R[i][1] + 1;
// Making it to 0-indexing
l--;
r--;
A[l]++;
if (r < N)
A[r]--;
}
// Iterating over the array
// to get the final array
for ( int i = 1; i < N; ++i) {
A[i] += A[i - 1];
}
// Variable to store the sum
int ans = 0;
// Hash to maintain previously
// occurred elements
unordered_set< int > s;
// Loop to find the maximum sum
for ( int i = 0; i < N; ++i) {
if (s.find(A[i]) == s.end())
ans += A[i];
s.insert(A[i]);
}
return ans;
} // Driver code int main()
{ int A[] = { 0, 0, 0, 0, 0, 0 };
int R[][2]
= { { 1, 3 }, { 4, 6 },
{ 3, 4 }, { 3, 3 } };
int N = sizeof (A) / sizeof (A[0]);
int M = sizeof (R) / sizeof (R[0]);
cout << uniqueSum(A, R, N, M);
return 0;
} |
// Java implementation to find the // sum of all unique elements of // the array after Q queries import java.util.*;
class GFG{
// Function to find the sum of // unique elements after Q Query static int uniqueSum( int A[], int R[][],
int N, int M)
{ // Updating the array after
// processing each query
for ( int i = 0 ; i < M; ++i)
{
int l = R[i][ 0 ], r = R[i][ 1 ] + 1 ;
// Making it to 0-indexing
l--;
r--;
A[l]++;
if (r < N)
A[r]--;
}
// Iterating over the array
// to get the final array
for ( int i = 1 ; i < N; ++i)
{
A[i] += A[i - 1 ];
}
// Variable to store the sum
int ans = 0 ;
// Hash to maintain previously
// occurred elements
HashSet<Integer> s = new HashSet<Integer>();
// Loop to find the maximum sum
for ( int i = 0 ; i < N; ++i)
{
if (!s.contains(A[i]))
ans += A[i];
s.add(A[i]);
}
return ans;
} // Driver code public static void main(String[] args)
{ int A[] = { 0 , 0 , 0 , 0 , 0 , 0 };
int R[][] = { { 1 , 3 }, { 4 , 6 },
{ 3 , 4 }, { 3 , 3 } };
int N = A.length;
int M = R.length;
System.out.print(uniqueSum(A, R, N, M));
} } // This code is contributed by gauravrajput1 |
# Python implementation to find the # sum of all unique elements of # the array after Q queries # Function to find the sum of # unique elements after Q Query def uniqueSum(A, R, N, M) :
# Updating the array after
# processing each query
for i in range ( 0 , M) :
l = R[i][ 0 ]
r = R[i][ 1 ] + 1
# Making it to 0-indexing
l - = 1
r - = 1
A[l] + = 1
if (r < N) :
A[r] - = 1
# Iterating over the array
# to get the final array
for i in range ( 1 , N) :
A[i] + = A[i - 1 ]
# Variable to store the sum
ans = 0
# Hash to maintain previously
# occurred elements
s = { chr }
# Loop to find the maximum sum
for i in range ( 0 , N) :
if (A[i] not in s) :
ans + = A[i]
s.add(A[i])
return ans
# Driver code A = [ 0 , 0 , 0 , 0 , 0 , 0 ]
R = [ [ 1 , 3 ], [ 4 , 6 ],
[ 3 , 4 ], [ 3 , 3 ] ]
N = len (A)
M = len (R)
print (uniqueSum(A, R, N, M))
# This code is contributed by Sanjit_Prasad |
// C# implementation to find the // sum of all unique elements of // the array after Q queries using System;
using System.Collections.Generic;
class GFG{
// Function to find the sum of // unique elements after Q Query static int uniqueSum( int []A, int [,]R,
int N, int M)
{ // Updating the array after
// processing each query
for ( int i = 0; i < M; ++i)
{
int l = R[i, 0], r = R[i, 1] + 1;
// Making it to 0-indexing
l--;
r--;
A[l]++;
if (r < N)
A[r]--;
}
// Iterating over the array
// to get the readonly array
for ( int i = 1; i < N; ++i)
{
A[i] += A[i - 1];
}
// Variable to store the sum
int ans = 0;
// Hash to maintain previously
// occurred elements
HashSet< int > s = new HashSet< int >();
// Loop to find the maximum sum
for ( int i = 0; i < N; ++i)
{
if (!s.Contains(A[i]))
ans += A[i];
s.Add(A[i]);
}
return ans;
} // Driver code public static void Main(String[] args)
{ int []A = { 0, 0, 0, 0, 0, 0 };
int [,]R = { { 1, 3 }, { 4, 6 },
{ 3, 4 }, { 3, 3 } };
int N = A.Length;
int M = R.GetLength(0);
Console.Write(uniqueSum(A, R, N, M));
} } // This code is contributed by Princi Singh |
<script> // Javascript implementation to find the // sum of all unique elements of // the array after Q queries // Function to find the sum of // unique elements after Q Query function uniqueSum(A, R, N, M)
{ // Updating the array after
// processing each query
for (let i = 0; i < M; ++i)
{
let l = R[i][0], r = R[i][1] + 1;
// Making it to 0-indexing
l--;
r--;
A[l]++;
if (r < N)
A[r]--;
}
// Iterating over the array
// to get the final array
for (let i = 1; i < N; ++i)
{
A[i] += A[i - 1];
}
// Variable to store the sum
let ans = 0;
// Hash to maintain previously
// occurred elements
let s = new Set();
// Loop to find the maximum sum
for (let i = 0; i < N; ++i)
{
if (!s.has(A[i]))
ans += A[i];
s.add(A[i]);
}
return ans;
} // Driver code let A = [ 0, 0, 0, 0, 0, 0 ];
let R = [[ 1, 3 ], [ 4, 6 ],
[ 3, 4 ], [ 3, 3 ]];
let N = A.length;
let M = R.length;
document.write(uniqueSum(A, R, N, M));
// This code is contributed by code_hunt. </script> |
6
Time Complexity: O(M + N)
Auxiliary Space: O(N)