Find frequency of the elements in given ranges

Given a 2-dimensional integer array arr[] representing N ranges, each of type [start_i, end_i] (start_i, end_i ≤ 10⁹) and Q queries represented in array query[], the task is to find the maximum occurrence of query[i] (query[i] ≤ 10⁹) in the given ranges for all i in the range [0, Q-1].

Examples:

Input: arr[] = {{1, 5}, {3, 6}, {5, 7}, {12, 15}}, query[] = {1, 3, 5, 13}
Output: {1, 2, 3, 1}
Explanation: The occurrence of 1 in the range is 1.
The occurrence of 3 in the range is 2.
The occurrence of 5 in the range is 3.
The occurrence of 13 in the range is 1.

Input: arr[] = {{2, 5}, {9, 11}, {3, 9}, {15, 20}}, query[] = {3, 9, 16, 55}
Output: {2, 2, 1, 0}

Naive approach: For each query, traverse the array and check whether query[j] is in the range of arr[i] = [start_i, end_i]. If it is in the range then increment the counter for the occurrence of that element and store this counter corresponding to query[j] in the result.

Time Complexity: O(Q * N)
Auxiliary Space: O(1)

Efficient approach: The problem can be solved based on the following idea:

Use the technique of prefix sum to store the occurrences of each element and it will help to find the answer for each query in constant time.

Follow the steps below to implement the above idea:

• Declare a one-dimensional array (say prefixSum[]) of length M (where M is the maximum element in arr[]) to store the occurrence of each element.
• Iterate over the array arr[] and do the following for each start_i and end_i :
  • Increment the value at prefixSum[start_i] by 1 
  • Decrement the value at prefixSum[end_i + 1] by 1
• Iterate over prefixSum[] array and calculate prefix sum. Using this, the occurrences of each element in the given ranges will get calculated 
• Iterate over the queries array and for each query get the value of prefixSum[query[i]] and store it into the resultant array.
• Return the result array.

 Below is the implementation of the above approach:

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;
#define MAX 10000000

// Function to find occurrences of
// each element in array
void findTheOccurrence(vector<vector<int> >& arr,
                      vector<int>& q,
                      vector<int>& result)
{
    vector<int> prefixSum(MAX);

    // Iterating over arr
    for (int i = 0; i < arr.size(); i++) {
        int start = arr[i][0];
        int end = arr[i][1];

        // Increment the value of
        // prefixSum at start
        prefixSum[start]++;

        // Decrement the value of
        // prefixSum at end + 1
        prefixSum[end + 1]--;
    }

    // Calculating the prefix sum
    for (int i = 1; i < prefixSum.size(); i++) {
        prefixSum[i]
            = prefixSum[i - 1] + prefixSum[i];
    }

    // Resolving each queries
    for (int i = 0; i < q.size(); i++) {
        int occurrence = prefixSum[q[i]];
        result.push_back(occurrence);
    }
}

// Driver code
int main()
{
    vector<vector<int> > arr
        = { { 1, 5 }, { 3, 6 }, { 5, 7 }, { 12, 15 } };
    vector<int> query = { 1, 3, 5, 13 };
    vector<int> result;

    // Function call
    findTheOccurrence(arr, query, result);

    for (auto i : result)
        cout << i << " ";
    return 0;
}

Java

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

class GFG
{
  static int MAX = 10000000;

  // Function to find occurrences of
  // each element in array
  static void findTheOccurrence(int[][] arr, int[] q,
                               ArrayList<Integer> result)
  {
    int[] prefixSum = new int[(MAX)];

    // Iterating over arr
    for (int i = 0; i < arr.length; i++) {
      int start = arr[i][0];
      int end = arr[i][1];

      // Increment the value of
      // prefixSum at start
      prefixSum[start]++;

      // Decrement the value of
      // prefixSum at end + 1
      prefixSum[end + 1]--;
    }

    // Calculating the prefix sum
    for (int i = 1; i < prefixSum.length; i++) {
      prefixSum[i]
        = prefixSum[i - 1] + prefixSum[i];
    }

    // Resolving each queries
    for (int i = 0; i < q.length; i++) {
      int occurrence = prefixSum[q[i]];
      result.add(occurrence);
    }
  }

  // Driver Code
  public static void main(String args[])
  {
    int[][] arr
      = { { 1, 5 }, { 3, 6 }, { 5, 7 }, { 12, 15 } };
    int[] query = { 1, 3, 5, 13 };
    ArrayList<Integer> result
      = new ArrayList<Integer>();


    // Function call
    findTheOccurrence(arr, query, result);

    for (int i : result)
      System.out.print(i + " ");
  }
}

// This code is contributed by sanjoy_62.

Python3

# Python3 code to implement the approach
MAX = 10000;

# Function to find occurrences of
# each element in array
def findTheOccurrence(arr, q, result):
    prefixSum = [0] * MAX

    # Iterating over arr
    for i in range(len(arr)):
        start = arr[i][0];
        end = arr[i][1];

        # Increment the value of
        # prefixSum at start
        prefixSum[start] += 1

        # Decrement the value of
        # prefixSum at end + 1
        prefixSum[end + 1] -= 1


    # Calculating the prefix sum
    for i in range(1, len(prefixSum)):
        prefixSum[i] = prefixSum[i - 1] + prefixSum[i];


    # Resolving each queries
    for i in range(0, len(q)):
        occurrence = prefixSum[q[i]];
        result.append(occurrence);
    return result;


# Driver code
arr  = [[ 1, 5 ], [ 3, 6 ], [ 5, 7 ], [ 12, 15 ] ];
query = [ 1, 3, 5, 13 ];
result = [];

# Function call
findTheOccurrence(arr, query, result);

for i in range(len(result)):
    print(result[i], end=" ");

# This code is contributed by Saurabh Jaiswal

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;

class GFG
{
 static int MAX = 10000000;

  // Function to find occurrences of
  // each element in array
  static void findTheOccurrence(int[,] arr, int[] q,
                               List<int> result)
  {
    int[] prefixSum = new int[(MAX)];

    // Iterating over arr
    for (int i = 0; i < arr.GetLength(0); i++) {
      int start = arr[i, 0];
      int end = arr[i, 1];

      // Increment the value of
      // prefixSum at start
      prefixSum[start]++;

      // Decrement the value of
      // prefixSum at end + 1
      prefixSum[end + 1]--;
    }

    // Calculating the prefix sum
    for (int i = 1; i < prefixSum.Length; i++) {
      prefixSum[i]
        = prefixSum[i - 1] + prefixSum[i];
    }

    // Resolving each queries
    for (int i = 0; i < q.Length; i++) {
      int occurrence = prefixSum[q[i]];
      result.Add(occurrence);
    }
  }

// Driver Code
public static void Main()
{
    int[,] arr
      = { { 1, 5 }, { 3, 6 }, { 5, 7 }, { 12, 15 } };
    int[] query = { 1, 3, 5, 13 };
    List<int> result
      = new List<int>();


    // Function call
    findTheOccurrence(arr, query, result);

    foreach (int i in result)
      Console.Write(i + " ");
}
}

// This code is contributed by avijitmondal998.

Javascript

<script>
// JS code to implement the approach
let MAX = 10000;

// Function to find occurrences of
// each element in array
function findTheOccurrence(arr, q, result)
{
    prefixSum = new Array(MAX).fill(0);

    // Iterating over arr
    for (var i = 0; i < arr.length; i++)
    {
        var start = arr[i][0];
        var end = arr[i][1];

        // Increment the value of
        // prefixSum at start
        prefixSum[start]++;

        // Decrement the value of
        // prefixSum at end + 1
        prefixSum[end + 1]--;
    }

    // Calculating the prefix sum
    for (var i = 1; i < prefixSum.length; i++) {
        prefixSum[i] = prefixSum[i - 1] + prefixSum[i];
    }

    // Resolving each queries
    for (var i = 0; i < q.length; i++) {
        var occurrence = prefixSum[q[i]];
        result.push(occurrence);
    }
    return result;
}

// Driver code
var arr  = [[ 1, 5 ], [ 3, 6 ], [ 5, 7 ], [ 12, 15 ] ];
var query = [ 1, 3, 5, 13 ];
var result = [];

// Function call
findTheOccurrence(arr, query, result);

for (var i = 0; i < result.length; i++)
    document.write(result[i] + " ");

// This code is contributed by phasing17
</script>

Output

1 2 3 1 

Time Complexity: O(M), where M is the maximum element of the array.
Auxiliary Space: O(MAX)

Note: This approach will not work here because of the given constraints

Space Optimized approach: The idea is similar to the above approach, but there will be the following changes:

We will use a map to store the frequency of elements in given ranges instead of creating a prefix sum array. If the range of elements in arr[] is high as 10⁹ then we can not create such a big length array. But in the map (ordered map) we can calculate the prefix sum while iterating over the map and can perform the rest of the operations as done in the above approach.

Follow the steps below to implement the above idea:

• Create a map (ordered map), where the key represents the element and the value represents the frequency of the key.
• Iterate over arr[]:
  • Increment the frequency of start_i by 1
  • Decrement the frequency of (end_i + 1) by 1
• Initialize two arrays to store the elements that appear in the map and the frequencies corresponding to that element.
• Iterate over the map and use prefix sum technique to calculate the frequency of each element and store them in the created arrays.
• Iterate over the query array and for each query:
  • Find the closest integer (say x) which just greater or equal to query[i] and is present in the map.
  • Check if x and query[i] are the same:
    • If they are the same then the corresponding frequency of x is the required frequency.
    • Otherwise, the frequency of the elements corresponding to the element just before x is the answer.
  • Store this frequency in the resultant array.
• Return the result array. 

Below is the implementation of the above approach:

C++

// C++ code to implement the approach

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

// Function to find the frequency
// of element in array arr
vector<int> findTheOccurrence(vector<vector<int> >& arr,
                             vector<int>& q,
                             vector<int>& result)
{
    int n = arr.size();

    // Map to store the elements
    // with their frequency
    map<long long, long long> mp;

    for (auto& i : arr) {
        mp[i[0]]++;
        mp[(long long)i[1] + 1]--;
    }

    vector<long long> range, freq;
    long long prefixSum = 0;

    for (auto i : mp) {

        // Calculate the frequency of element
        // using prefix sum technique.
        prefixSum = prefixSum
                    + (long long)i.second;

        // Store the element of arr in range
        range.push_back(i.first);

        // Store its corresponding frequency
        freq.push_back(prefixSum);
    }

    // Iterate over the query
    for (auto p : q) {

        // Find the lower_bound of query p
        int idx = lower_bound(range.begin(),
                              range.end(), p)
                  - range.begin();

        // If the lower_bound doesn't exist
        if (idx >= 0 && idx < range.size()) {

            // Check if element
            // which we are searching
            // exists in the range or not.
            // If it doesn't exist then
            // lower bound will return
            // the next element which is
            // just greater than p
            if (range[idx] != p)
                idx--;
            if (idx >= 0)
                result.push_back(freq[idx]);

            // If no such element exist
            else
                result.push_back(0);
        }

        // If idx is range size,
        // it means all elements
        // are smaller than p.
        // So next smaller element will be
        // at range.size() - 1
        else {
            result.push_back(freq[range.size()
                                  - 1]);
        }
    }

    // Return the final result
    return result;
}

// Driver code
int main()
{
    vector<vector<int> > arr
        = { { 1, 5 }, { 3, 6 }, { 5, 7 }, { 12, 15 } };
    vector<int> q = { 1, 3, 5, 13 };
    vector<int> result;

    // Function call
    findTheOccurrence(arr, q, result);

    for (auto i : result)
        cout << i << " ";
    return 0;
}

Java

// Java code to implement the approach
import java.util.*;

class GFG
{

  // This function returns the position
  // of the leftmost array element that is
  // greater than or equal to the key element
  static int lower_bound(ArrayList<Integer> arr, int ele)
  {
    for (var i = 0; i < arr.size(); i++) {
      if (arr.get(i) >= ele)
        return i;
    }
    return arr.size();
  }

  // Function to find the frequency
  // of element in array arr
  static ArrayList<Integer>
    findTheOccurrence(int[][] arr, int[] q,
                      ArrayList<Integer> result)
  {
    int n = arr.length;

    // Map to store the elements
    // with their frequency
    TreeMap<Integer, Integer> mp
      = new TreeMap<Integer, Integer>();

    for (int i = 0; i < arr.length; i++) {
      if (!mp.containsKey(arr[i][0]))
        mp.put(arr[i][0], 0);
      if (!mp.containsKey(arr[i][1] + 1))
        mp.put(arr[i][1] + 1, 0);
      mp.put(arr[i][0], mp.get(arr[i][0]) + 1);
      mp.put(arr[i][1] + 1,
             mp.get(arr[i][1] + 1) - 1);
    }

    ArrayList<Integer> range = new ArrayList<Integer>();
    ArrayList<Integer> freq = new ArrayList<Integer>();
    int prefixSum = 0;

    for (Map.Entry<Integer, Integer> i :
         mp.entrySet()) {

      // Calculate the frequency of element
      // using prefix sum technique.
      prefixSum = prefixSum + i.getValue();

      // Store the element of arr in range
      range.add(i.getKey());

      // Store its corresponding frequency
      freq.add(prefixSum);
    }

    // Iterate over the query
    for (Integer p : q) {

      // Find the lower_bound of query p
      int idx = lower_bound(range, p);

      // If the lower_bound doesn't exist
      if (idx >= 0 && idx < range.size()) {

        // Check if element
        // which we are searching
        // exists in the range or not.
        // If it doesn't exist then
        // lower bound will return
        // the next element which is
        // just greater than p
        if (range.get(idx) != p)
          idx--;
        if (idx >= 0)
          result.add(freq.get(idx));

        // If no such element exist
        else
          result.add(0);
      }

      // If idx is range size,
      // it means all elements
      // are smaller than p.
      // So next smaller element will be
      // at range.size() - 1
      else {
        result.add(freq.get(range.size() - 1));
      }
    }

    // Return the final result
    return result;
  }

  // Driver code
  public static void main(String[] args)
  {
    int[][] arr
      = { { 1, 5 }, { 3, 6 }, { 5, 7 }, { 12, 15 } };
    int[] q = { 1, 3, 5, 13 };
    ArrayList<Integer> result
      = new ArrayList<Integer>();

    // Function call
    findTheOccurrence(arr, q, result);

    for (Integer i : result)
      System.out.print(i + " ");
  }
}

// This code is contributed by phasing17

Python3

# Python3 code to implement the approach
import bisect

# Function to find the frequency
# of element in array arr
def findTheOccurrence(arr, q, result):
    n = len(arr)

    # Map to store the elements
    # with their frequency
    mp = {}

    for i in arr:
        if i[0] not in mp:
            mp[i[0]] = 0
        if (i[1] + 1) not in mp:
            mp[i[1] + 1] = 0
        mp[i[0]] += 1
        mp[i[1] + 1] -= 1


    range = []
    freq = []
    prefixSum = 0

    for i in sorted(mp.keys()):

        # Calculate the frequency of element
        # using prefix sum technique.
        prefixSum = prefixSum + mp[i]

        # Store the element of arr in range
        range.append(i)

        # Store its corresponding frequency
        freq.append(prefixSum)


    # Iterate over the query
    for p in q:

        # Find the lower_bound of query p
        idx = bisect.bisect_left(range, p)

        # If the lower_bound doesn't exist
        if (idx >= 0 and idx < len(range)):

            # Check if element
            # which we are searching
            # exists in the range or not.
            # If it doesn't exist then
            # lower bound will return
            # the next element which is
            # just greater than p
            if (range[idx] != p):
                idx -= 1
            if (idx >= 0):
                result.append(freq[idx])

            # If no such element exist
            else:
                result.append(0)

        # If idx is range size,
        # it means all elements
        # are smaller than p.
        # So next smaller element will be
        # at range.size() - 1
        else :
            result.append(freq[len(range) - 1])

    # Return the final result
    return result



# Driver code
arr = [[ 1, 5 ], [ 3, 6 ], [ 5, 7 ], [ 12, 15 ] ]
q = [ 1, 3, 5, 13 ]
result = []

# Function call
result = findTheOccurrence(arr, q, result);

print(" ".join(list(map(str, result))))



# This code is contributed by phasing17

C#

// C# code to implement the approach
using System;
using System.Collections.Generic;

class GFG {

  // This function returns the position
  // of the leftmost array element that is
  // greater than or equal to the key element
  static int lower_bound(List<int> arr, int ele)
  {
    for (var i = 0; i < arr.Count; i++) {
      if (arr[i] >= ele)
        return i;
    }
    return arr.Count;
  }

  // Function to find the frequency
  // of element in array arr
  static List<int> findTheOccurrence(int[, ] arr, int[] q,
                                     List<int> result)
  {

    // Map to store the elements
    // with their frequency
    Dictionary<int, int> mp
      = new Dictionary<int, int>();

    for (int i = 0; i < arr.GetLength(0); i++) {
      if (!mp.ContainsKey(arr[i, 0]))
        mp[arr[i, 0]] = 0;
      if (!mp.ContainsKey(arr[i, 1] + 1))
        mp[arr[i, 1] + 1] = 0;
      mp[arr[i, 0]] += 1;
      mp[arr[i, 1] + 1] -= 1;
    }

    List<int> range = new List<int>();
    List<int> freq = new List<int>();
    int prefixSum = 0;

    var mpKeys = new List<int>(mp.Keys);
    mpKeys.Sort();

    foreach(var i in mpKeys)
    {

      // Calculate the frequency of element
      // using prefix sum technique.
      prefixSum = prefixSum + mp[i];

      // Store the element of arr in range
      range.Add(i);

      // Store its corresponding frequency
      freq.Add(prefixSum);
    }

    // Iterate over the query
    foreach(var p in q)
    {

      // Find the lower_bound of query p
      int idx = lower_bound(range, p);

      // If the lower_bound doesn't exist
      if (idx >= 0 && idx < range.Count) {

        // Check if element
        // which we are searching
        // exists in the range or not.
        // If it doesn't exist then
        // lower bound will return
        // the next element which is
        // just greater than p
        if (range[idx] != p)
          idx--;
        if (idx >= 0)
          result.Add(freq[idx]);

        // If no such element exist
        else
          result.Add(0);
      }

      // If idx is range size,
      // it means all elements
      // are smaller than p.
      // So next smaller element will be
      // at range.size() - 1
      else {
        result.Add(freq[range.Count - 1]);
      }
    }

    // Return the final result
    return result;
  }

  // Driver code
  public static void Main(string[] args)
  {
    int[, ] arr
      = { { 1, 5 }, { 3, 6 }, { 5, 7 }, { 12, 15 } };
    int[] q = { 1, 3, 5, 13 };
    List<int> result = new List<int>();

    // Function call
    findTheOccurrence(arr, q, result);

    foreach(var i in result) Console.Write(i + " ");
  }
}

// This code is contributed by phasing17

Javascript

// JavaScript code to implement the approach


//This function returns the position
//of the leftmost array element that is
//greater than or equal to the key element
function lower_bound(arr, ele)
{
    for (var i = 0; i < arr.length; i++)
    {
        if (arr[i] >= ele)
            return i;
    }
    return arr.length - 1;
}

// Function to find the frequency
// of element in array arr
function findTheOccurrence(arr, q, result)
{
    let n = arr.length;

    // Map to store the elements
    // with their frequency
    let mp = {};

    for (var i of arr) {
        if (!mp.hasOwnProperty(i[0]))
            mp[i[0]] = 0;
        if (!mp.hasOwnProperty(i[1] + 1))
            mp[i[1] + 1] = 0;
        mp[i[0]]++;
        mp[i[1] + 1]--;
    }

    let range = [];
    let freq = [];
    let prefixSum = 0;

    for (let [i, val] of Object.entries(mp)) {

        // Calculate the frequency of element
        // using prefix sum technique.
        prefixSum = prefixSum + val;

        // Store the element of arr in range
        range.push(parseInt(i));

        // Store its corresponding frequency
        freq.push(prefixSum);
    }

    // Iterate over the query
    for (var p of q) {

        // Find the lower_bound of query p
        let idx = lower_bound(range, p);

        // If the lower_bound doesn't exist
        if (idx >= 0 && idx < range.length) {

            // Check if element
            // which we are searching
            // exists in the range or not.
            // If it doesn't exist then
            // lower bound will return
            // the next element which is
            // just greater than p
            if (range[idx] != p)
                idx--;
            if (idx >= 0)
                result.push(freq[idx]);

            // If no such element exist
            else
                result.push(0);
        }

        // If idx is range size,
        // it means all elements
        // are smaller than p.
        // So next smaller element will be
        // at range.size() - 1
        else {
            result.push(freq[range.length - 1]);
        }
    }

    // Return the final result
    return result;
}

// Driver code
let arr = [[ 1, 5 ], [ 3, 6 ], [ 5, 7 ], [ 12, 15 ] ];
let q = [ 1, 3, 5, 13 ];
let result = [];

// Function call
result = findTheOccurrence(arr, q, result);

for (var i of result)
    process.stdout.write(i + " ");



//This code is contributed by phasing17

Output

1 2 3 1 

Time Complexity: O(N * log(N))
Auxiliary Space: O(N)