Count of numbers appearing in the given ranges at-least K times
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++
#include <bits/stdc++.h>
using namespace std;
int countNumbers( int n, int k, int rangeLvalue[],
int rangeRvalue[])
{
int count = 0;
map< int , int > freq;
for ( int i = 0; i < n; i++) {
for ( int j = rangeLvalue[i]; j <= rangeRvalue[i]; j++)
{
if (freq.find(j) == freq.end())
freq.insert(make_pair(j, 1));
else
freq[j]++;
}
}
for ( auto itr = freq.begin(); itr != freq.end(); itr++)
{
if ((*itr).second >= k) {
count++;
}
}
return count;
}
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
import java.io.*;
import java.util.*;
class GFG
{
static int countNumbers( int n, int k, int [] rangeLvalue,
int [] rangeRvalue)
{
int count = 0 ;
HashMap<Integer, Integer> freq = new HashMap<>();
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 );
}
}
for (HashMap.Entry<Integer, Integer> entry : freq.entrySet())
{
if (entry.getValue() >= k)
count++;
}
return count;
}
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));
}
}
|
Python3
def countNumbers(n, k, rangeLvalue,
rangeRvalue):
count = 0
freq = dict ()
for i in range (n):
for j in range (rangeLvalue[i],
rangeRvalue[i] + 1 ):
freq[j] = freq.get(j, 0 ) + 1
for itr in freq:
if (freq[itr] > = k):
count + = 1
return count
n, k = 3 , 2
rangeLvalue = [ 91 , 92 , 97 ]
rangeRvalue = [ 94 , 97 , 99 ]
print (countNumbers(n, k, rangeLvalue,
rangeRvalue))
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int countNumbers( int n, int k, int [] rangeLvalue,
int [] rangeRvalue)
{
int count = 0;
Dictionary< int , int > freq = new Dictionary< int , int >();
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;
}
}
foreach (KeyValuePair< int , int > entry in freq)
{
if (entry.Value >= k)
count++;
}
return count;
}
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));
}
}
|
Javascript
<script>
function countNumbers(n,k,rangeLvalue,rangeRvalue)
{
let count = 0;
let freq = new Map();
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);
}
}
for (let [key, value] of freq.entries())
{
if (value >= k)
count++;
}
return count;
}
let n = 3, k = 2;
let rangeLvalue=[91, 92, 97];
let rangeRvalue =[94, 97, 99];
document.write(countNumbers(n, k, rangeLvalue,
rangeRvalue));
</script>
|
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++
#include <bits/stdc++.h>
using namespace std;
int countNumbers( int n, int k, int rangeLvalue[],
int rangeRvalue[])
{
int count = 0;
int maxn = INT_MIN;
for ( int i = 0; i < n; i++)
if (rangeRvalue[i] > maxn)
maxn = rangeRvalue[i];
int preSum[maxn + 5] = { 0 };
for ( int i = 0; i < n; i++) {
preSum[rangeLvalue[i]]++;
preSum[rangeRvalue[i] + 1]--;
}
for ( int i = 1; i <= maxn; i++) {
preSum[i] += preSum[i - 1];
}
for ( int i = 1; i <= maxn; i++) {
if (preSum[i] >= k) {
count++;
}
}
return count;
}
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
class Geeks {
static int countNumbers( int n, int k, int rangeLvalue[],
int rangeRvalue[])
{
int count = 0 ;
int maxn = Integer.MIN_VALUE;
for ( int i = 0 ; i < n; i++)
if (rangeRvalue[i] > maxn)
maxn = rangeRvalue[i];
int preSum[] = new int [maxn + 5 ];
for ( int i = 0 ; i < (maxn + 5 ); i++)
preSum[i] = 0 ;
for ( int i = 0 ; i < n; i++)
{
preSum[rangeLvalue[i]]++;
preSum[rangeRvalue[i] + 1 ]--;
}
for ( int i = 1 ; i <= maxn; i++) {
preSum[i] += preSum[i - 1 ];
}
for ( int i = 1 ; i <= maxn; i++) {
if (preSum[i] >= k) {
count++;
}
}
return count;
}
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));
}
}
|
Python
def countNumbers(n, k, rangeLvalue, rangeRvalue):
count = 0
maxn = - float ( 'inf' )
for i in range (n):
if rangeRvalue[i] > maxn:
maxn = rangeRvalue[i]
preSum = [ 0 ] * (maxn + 5 )
for i in range (n):
preSum[rangeLvalue[i]] + = 1
preSum[rangeRvalue[i] + 1 ] - = 1
for i in range ( 1 , maxn + 1 ):
preSum[i] + = preSum[i - 1 ]
for i in range ( 1 , maxn + 1 ):
if preSum[i] > = k:
count + = 1
return count
n = 3
k = 2
rangeLvalue = [ 91 , 92 , 97 ]
rangeRvalue = [ 94 , 97 , 99 ]
print (countNumbers(n, k, rangeLvalue, rangeRvalue))
|
C#
using System;
class GFG
{
static int countNumbers( int n, int k,
int []rangeLvalue,
int []rangeRvalue)
{
int count = 0;
int maxn = Int32.MinValue;
for ( int i = 0; i < n; i++)
if (rangeRvalue[i] > maxn)
maxn = rangeRvalue[i];
int []preSum = new int [maxn + 5];
for ( int i = 0; i < (maxn + 5); i++)
preSum[i] = 0;
for ( int i = 0; i < n; i++)
{
preSum[rangeLvalue[i]]++;
preSum[rangeRvalue[i] + 1]--;
}
for ( int i = 1; i <= maxn; i++)
{
preSum[i] += preSum[i - 1];
}
for ( int i = 1; i <= maxn; i++)
{
if (preSum[i] >= k)
{
count++;
}
}
return count;
}
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));
}
}
|
Javascript
<script>
function countNumbers(n, k, rangeLvalue, rangeRvalue)
{
var count = 0;
var maxn = -1000000000;
for ( var i = 0; i < n; i++)
if (rangeRvalue[i] > maxn)
maxn = rangeRvalue[i];
var preSum = Array(maxn +5).fill(0);
for ( var i = 0; i < n; i++) {
preSum[rangeLvalue[i]]++;
preSum[rangeRvalue[i] + 1]--;
}
for ( var i = 1; i <= maxn; i++) {
preSum[i] += preSum[i - 1];
}
for ( var i = 1; i <= maxn; i++) {
if (preSum[i] >= k) {
count++;
}
}
return count;
}
var n = 3, k = 2;
var rangeLvalue = [91, 92, 97];
var rangeRvalue = [94, 97, 99];
document.write( countNumbers(n, k, rangeLvalue, rangeRvalue));
</script>
|
Time-Complexity: O(N + max(rangeRvalue[]))
Auxiliary-Space: O(N)
Last Updated :
03 Jun, 2021
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