Given N ranges and a number K, the task is to find the total count of numbers that appear 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 occurred 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++ brute-force program to find count of // numbers appearing in the given // ranges at-least K times #include <bits/stdc++.h> 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< int , int > 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;
} |
// Java brute-force program to find count of // numbers appearing in the given // ranges at-least K times import java.io.*;
import java.util.*;
class GFG
{ // Function to find the no of occurrence
static int countNumbers( int n, int k, int [] rangeLvalue,
int [] rangeRvalue)
{
int count = 0 ;
// Map to store frequency of elements
// in the range
HashMap<Integer, Integer> freq = new HashMap<>();
// increment the value of the elements
// in all of the ranges
for ( int i = 0 ; i < n; i++)
{
for ( int j = rangeLvalue[i]; j <= rangeRvalue[i]; j++)
{
if (!freq.containsKey(j))
freq.put(j, 1 );
else
freq.put(j, freq.get(j) + 1 );
}
}
// Traverse the map to check the frequency
// of numbers greater than equals to k
for (HashMap.Entry<Integer, Integer> entry : freq.entrySet())
{
// check if a number appears atleast k times
if (entry.getValue() >= 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 // sanjeev2552 |
# 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 |
// C# brute-force program to find count of // numbers appearing in the given // ranges at-least K times using System;
using System.Collections.Generic;
class GFG
{ // Function to find the no of occurrence
static int countNumbers( int n, int k, int [] rangeLvalue,
int [] rangeRvalue)
{
int count = 0;
// Map to store frequency of elements
// in the range
Dictionary< int , int > freq = new Dictionary< int , int >();
// increment the value of the elements
// in all of the ranges
for ( int i = 0; i < n; i++)
{
for ( int j = rangeLvalue[i]; j <= rangeRvalue[i]; j++)
{
if (!freq.ContainsKey(j))
freq.Add(j, 1);
else
freq[j] = freq[j] + 1;
}
}
// Traverse the map to check the frequency
// of numbers greater than equals to k
foreach (KeyValuePair< int , int > entry in freq)
{
// check if a number appears atleast k times
if (entry.Value >= 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 Rajput-Ji |
<script> // JavaScript brute-force program to find count of // numbers appearing in the given // ranges at-least K times // Function to find the no of occurrence
function countNumbers(n,k,rangeLvalue,rangeRvalue)
{
let count = 0;
// Map to store frequency of elements
// in the range
let freq = new Map();
// increment the value of the elements
// in all of the ranges
for (let i = 0; i < n; i++)
{
for (let j = rangeLvalue[i];
j <= rangeRvalue[i]; j++)
{
if (!freq.has(j))
freq.set(j, 1);
else
freq.set(j, freq.get(j) + 1);
}
}
// Traverse the map to check the frequency
// of numbers greater than equals to k
for (let [key, value] of freq.entries())
{
// check if a number appears atleast k times
if (value >= k)
// increase the counter
// if condition satisfies
count++;
}
// return the result
return count;
}
// Driver Code
let n = 3, k = 2;
let rangeLvalue=[91, 92, 97];
let rangeRvalue =[94, 97, 99];
document.write(countNumbers(n, k, rangeLvalue,
rangeRvalue));
// This code is contributed by unknown2108 </script> |
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 occurred in the given range on doing a pre-sum.
Below is the implementation of the above approach.
// C++ efficient program to find count of // numbers appearing in the given // ranges at-least K times #include <bits/stdc++.h> 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 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 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 = [ 0 ] * (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# 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 |
<script> // Javascript efficient program to find count of // numbers appearing in the given // ranges at-least K times // Function to find the no of occurrence function countNumbers(n, k, rangeLvalue, rangeRvalue)
{ var count = 0;
// maximum value of the range
var maxn = -1000000000;
for ( var i = 0; i < n; i++)
if (rangeRvalue[i] > maxn)
maxn = rangeRvalue[i];
// counter array
var preSum = Array(maxn +5).fill(0);
// incrementing and decrementing the
// leftmost and next value of rightmost value
// of each range by 1 respectively
for ( var i = 0; i < n; i++) {
preSum[rangeLvalue[i]]++;
preSum[rangeRvalue[i] + 1]--;
}
// presum gives the no of occurrence of
// each element
for ( var i = 1; i <= maxn; i++) {
preSum[i] += preSum[i - 1];
}
for ( var 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 var n = 3, k = 2;
var rangeLvalue = [91, 92, 97];
var rangeRvalue = [94, 97, 99];
document.write( countNumbers(n, k, rangeLvalue, rangeRvalue)); </script> |
4
Time-Complexity: O(N + max(rangeRvalue[]))
Auxiliary-Space: O(N)