Count of numbers appearing in the given ranges at-least K times

Given N ranges and a number K, the task is to find total count of numbers that appears at least K times in the given ranges.

Examples:

Input:
N = 3, K = 2
Range 1: [91, 94]
Range 2: [92, 97]
Range 3: [97, 99]
Output : 4

Explanation: Ranges are 91 to 94, 92 to 97, 97 to 99 and the numbers that occured atleast 2 times are 92, 93, 94, 97.

Input :
N = 2, K = 3
Range 1 = [1, 4]
Range 2 = [5, 9]
Output : 0
Explanation: No element occurred 3 times in the given ranges.

Naive Approach: A naive approach would be to traverse each range and increase the count for each element, finally check whether the count of each element suffices the required value.

Below is the implementation of the above approach:

C++

 // C++ brute-force program to find count of // numbers appearing in the given // ranges at-least K times #include using namespace std;    // Function to find the no of occurrence int countNumbers(int n, int k, int rangeLvalue[],                               int rangeRvalue[]) {     int count = 0;        // Map to store frequency of elements     // in the range     map freq;        for (int i = 0; i < n; i++) {            // increment the value of the elements         // in all of the ranges         for (int j = rangeLvalue[i]; j <= rangeRvalue[i]; j++)         {             if (freq.find(j) == freq.end())                 freq.insert(make_pair(j, 1));             else                 freq[j]++;         }     }        // Traverse the map to check the frequency     // of numbers greater than equals to k     for (auto itr = freq.begin(); itr != freq.end(); itr++)     {            // check if a number appears atleast k times         if ((*itr).second >= k) {                // increase the counter             // if condition satisfies             count++;         }     }        // return the result     return count; }    // Driver Code int main() {     int n = 3, k = 2;     int rangeLvalue[] = { 91, 92, 97 };     int rangeRvalue[] = { 94, 97, 99 };     cout << countNumbers(n, k, rangeLvalue, rangeRvalue);     return 0; }

Python3

 # Python3 brute-force program to find  # count of numbers appearing in the  # given ranges at-least K times    # Function to find the no of occurrence def countNumbers(n, k, rangeLvalue,                             rangeRvalue):        count = 0        # Map to store frequency of elements     # in the range     freq = dict()        for i in range(n):            # increment the value of the elements         # in all of the ranges         for j in range(rangeLvalue[i],                         rangeRvalue[i] + 1):             freq[j] = freq.get(j, 0) + 1            # Traverse the map to check the frequency     # of numbers greater than equals to k     for itr in freq:            # check if a number appears          # atleast k times         if (freq[itr] >= k):                # increase the counter             # if condition satisfies             count += 1                # return the result     return count    # Driver Code n, k = 3, 2 rangeLvalue = [91, 92, 97] rangeRvalue = [94, 97, 99] print(countNumbers(n, k, rangeLvalue,                          rangeRvalue))    # This code is contributed by mohit kumar

Output:

4

Efficient Solution: A better solution is to keep track of the ranges by incrementing the value of the leftmost element of the range and decrementing the next element of the rightmost element of the given range in the counter array. Do this for all ranges. This is done as it gives the idea of how many times a number occured in the given range on doing a pre-sum.

Below is the implementation of the above approach.

C++

 // C++ efficient program to find count of // numbers appearing in the given // ranges at-least K times #include using namespace std;    // Function to find the no of occurrence int countNumbers(int n, int k, int rangeLvalue[],                               int rangeRvalue[]) {     int count = 0;        // maximum value of the range     int maxn = INT_MIN;     for (int i = 0; i < n; i++)         if (rangeRvalue[i] > maxn)             maxn = rangeRvalue[i];        // counter array     int preSum[maxn + 5] = { 0 };        // incrementing and decrementing the     // leftmost and next value of rightmost value     // of each range by 1 respectively     for (int i = 0; i < n; i++) {         preSum[rangeLvalue[i]]++;         preSum[rangeRvalue[i] + 1]--;     }        // presum gives the no of occurrence of     // each element     for (int i = 1; i <= maxn; i++) {         preSum[i] += preSum[i - 1];     }        for (int i = 1; i <= maxn; i++) {            // check if the number appears atleast k times         if (preSum[i] >= k) {                // increase the counter if             // condition satisfies             count++;         }     }        // return the result     return count; }    // Driver Code int main() {     int n = 3, k = 2;        int rangeLvalue[] = { 91, 92, 97 };     int rangeRvalue[] = { 94, 97, 99 };        cout << countNumbers(n, k, rangeLvalue, rangeRvalue);        return 0; }

Java

 // Java efficient program to find count of // numbers appearing in the given // ranges at-least K times class Geeks {    // Function to find the no of occurrence static int countNumbers(int n, int k, int rangeLvalue[],                                      int rangeRvalue[]) {     int count = 0;        // maximum value of the range     int maxn = Integer.MIN_VALUE;     for (int i = 0; i < n; i++)         if (rangeRvalue[i] > maxn)             maxn = rangeRvalue[i];        // counter array     int preSum[] = new int[maxn + 5];     for(int i = 0; i < (maxn + 5); i++)     preSum[i] = 0;        // incrementing and decrementing the     // leftmost and next value of rightmost value     // of each range by 1 respectively     for (int i = 0; i < n; i++)     {         preSum[rangeLvalue[i]]++;         preSum[rangeRvalue[i] + 1]--;     }        // presum gives the no of occurrence of     // each element     for (int i = 1; i <= maxn; i++) {         preSum[i] += preSum[i - 1];     }        for (int i = 1; i <= maxn; i++) {            // check if the number appears atleast k times         if (preSum[i] >= k) {                // increase the counter if             // condition satisfies             count++;         }     }        // return the result     return count; }    // Driver Code public static void main(String args[]) {     int n = 3, k = 2;        int rangeLvalue[] = { 91, 92, 97 };     int rangeRvalue[] = { 94, 97, 99 };        System.out.println(countNumbers(n, k, rangeLvalue,                                         rangeRvalue));    } }    // This code is contributed by ankita_saini

Python

 # Python efficient program to find count of # numbers appearing in the given # ranges at-least K times    # Function to find the no of occurrence def countNumbers(n, k, rangeLvalue, rangeRvalue):     count = 0        # maximum value of the range     maxn = -float('inf')     for i in range(n):         if rangeRvalue[i] > maxn:             maxn = rangeRvalue[i]        # counter array     preSum = *(maxn + 5)        # incrementing and decrementing the     # leftmost and next value of rightmost value     # of each range by 1 respectively     for i in range(n):         preSum[rangeLvalue[i]] += 1         preSum[rangeRvalue[i] + 1] -= 1        # presum gives the no of occurrence of     # each element     for i in range(1, maxn+1):         preSum[i] += preSum[i - 1]        for i in range(1, maxn+1):            # check if the number appears atleast k times         if preSum[i] >= k:                # increase the counter if             # condition satisfies             count += 1        # return the result     return count       # Driver Code n = 3 k = 2    rangeLvalue = [91, 92, 97] rangeRvalue = [94, 97, 99]    print(countNumbers(n, k, rangeLvalue, rangeRvalue))    # This code is contributed by ankush_953

C#

 // C# efficient program to  // find count of numbers  // appearing in the given // ranges at-least K times using System;    class GFG {    // Function to find the  // no of occurrence static int countNumbers(int n, int k,                          int []rangeLvalue,                         int []rangeRvalue) {     int count = 0;        // maximum value of the range     int maxn = Int32.MinValue;     for (int i = 0; i < n; i++)         if (rangeRvalue[i] > maxn)             maxn = rangeRvalue[i];        // counter array     int []preSum = new int[maxn + 5];     for(int i = 0; i < (maxn + 5); i++)     preSum[i] = 0;        // incrementing and decrementing      // the leftmost and next value      // of rightmost value of each      // range by 1 respectively     for (int i = 0; i < n; i++)     {         preSum[rangeLvalue[i]]++;         preSum[rangeRvalue[i] + 1]--;     }        // presum gives the no of      // occurrence of each element     for (int i = 1; i <= maxn; i++)      {         preSum[i] += preSum[i - 1];     }        for (int i = 1; i <= maxn; i++)     {            // check if the number          // appears atleast k times         if (preSum[i] >= k)          {                // increase the counter if             // condition satisfies             count++;         }     }        // return the result     return count; }    // Driver Code public static void Main(String []args) {     int n = 3, k = 2;        int []rangeLvalue = { 91, 92, 97 };     int []rangeRvalue = { 94, 97, 99 };        Console.WriteLine(countNumbers(n, k, rangeLvalue,                                         rangeRvalue)); } }    // This code is contributed // by ankita_saini

Output:

4

Time-Complexity: O(N + max(rangeRvalue[]))
Auxillary-Space: O(N)

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