Minimize range [L, R] to divide Array into K subarrays with majority elements in [L, R]
Last Updated :
11 Mar, 2022
Given an array arr[] of size N, the task is to find the minimum value range [L, R] such that:
- The array can be divided into K sub-arrays.
- The elements within the range [L, R] are greater than the elements which are out of the range[l, r].
Examples:
Input: arr[] = {1, 2, 2, 2}, K = 2
Output: 2 2
Explanation: [2, 2] is the range with minimum distance which can split the array into two sub-arrays as
{1, 2, 2} -> inrange = 2, outrange =1
{2} -> inrange =1, outrange = 0.
In the 2 splits number of elements inrange > outrange.
Input: arr[] = {1, 2, 3, 4}, K = 3
Output: 1 4
Explanation : [1, 4] is the range with minimum distance because the array can be splitted into 3 subarrays as
{1, 2} -> inrange = 2, outrange =1 ,
{3} -> inrange =1, outrange = 0,
{4} -> inrange = 1, outrange = 0.
In the 3 splits number of elements inrange > outrange.
Naive Approach: This can be done by checking the ranges of every size from 1 to the size of the array, then checking the number of elements in the range and the number of elements out of the range, later checking if their difference is greater than or equal to K.
Time Complexity: O(N2)
Auxiliary Space: O(N)
Efficient approach: This approach is based on a Binary search of the range [1, N] with Hashing technique and prefix sum to keep track of the number of elements in the array that are in the range till i and checking if there is a possibility for any range which can divide the array into K sub-arrays that follow the given condition. Follow the mentioned steps:
- Initialize the count vector to store the frequencies of the elements of the array.
- Initialize a vector prefix sum to store the number of elements till that index.
- Now traverse through the array from[1, N] and store the count of elements till i using the prefix sum technique.
- Initialize can= 0, and l, r to store the range and low = 0, high = N to perform a binary search on the size of the range.
- Perform binary search while low ≤ high.
- Initialize mid as (low + high)/2.
- Iterate using the for loop from [1, N-mid+1] using i to check which range of size mid.
- Now Count the number of elements in the range [i, i+mid-1] and out of the range.
- Check if the difference of the elements in range and out range is greater than or equal to K.
- Store the range in the l, r and make can =1 if it is greater than equal to K.
- If there is a possibility then make high as mid-1
- Else low= mid +1.
- Print the range.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
void find_minrange(vector< int > arr, int n, int k)
{
vector< int > count(n + 1);
for ( auto x : arr)
count[x]++;
vector< int > prefixsum(n + 1);
for ( int i = 1; i <= n; i++)
prefixsum[i] = prefixsum[i - 1]
+ count[i];
int low = 1, high = n;
int l, r;
while (low <= high) {
int mid = (low + high) / 2;
bool can = false ;
for ( int i = 1; i <= n - mid + 1; i++) {
int inrange
= prefixsum[i + mid - 1]
- prefixsum[i - 1];
int outrange = n - inrange;
if (inrange - outrange >= k) {
can = true ;
l = i;
r = i + mid - 1;
break ;
}
}
if (can) {
high = mid - 1;
}
else {
low = mid + 1;
}
}
cout << l << " " << r;
}
int main()
{
vector< int > arr = { 1, 2, 2, 2 };
int K = 2;
int N = arr.size();
find_minrange(arr, N, K);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static void find_minrange( int [] arr, int n, int k)
{
int [] count = new int [n + 1 ];
for ( int i = 0 ; i < arr.length; i++)
count[arr[i]]++;
int [] prefixsum = new int [n + 1 ];
for ( int i = 1 ; i <= n; i++)
prefixsum[i] = prefixsum[i - 1 ] + count[i];
int low = 1 , high = n;
int l = 0 , r = 0 ;
while (low <= high) {
int mid = (low + high) / 2 ;
boolean can = false ;
for ( int i = 1 ; i <= n - mid + 1 ; i++) {
int inrange = prefixsum[i + mid - 1 ]
- prefixsum[i - 1 ];
int outrange = n - inrange;
if (inrange - outrange >= k) {
can = true ;
l = i;
r = i + mid - 1 ;
break ;
}
}
if (can == true ) {
high = mid - 1 ;
}
else {
low = mid + 1 ;
}
}
System.out.print(l + " " + r);
}
public static void main(String args[])
{
int [] arr = { 1 , 2 , 2 , 2 };
int K = 2 ;
int N = arr.length;
find_minrange(arr, N, K);
}
}
|
Python3
def find_minrange(arr, n, k):
count = [ 0 for _ in range (n + 1 )]
for x in arr:
count[x] + = 1
prefixsum = [ 0 for _ in range (n + 1 )]
for i in range ( 1 , n + 1 ):
prefixsum[i] = prefixsum[i - 1 ] + count[i]
low, high = 1 , n
l, r = 0 , 0
while (low < = high):
mid = (low + high) / / 2
can = False
for i in range ( 1 , n - mid + 1 + 1 ):
inrange = prefixsum[i + mid - 1 ] - prefixsum[i - 1 ]
outrange = n - inrange
if (inrange - outrange > = k):
can = True
l = i
r = i + mid - 1
break
if (can):
high = mid - 1
else :
low = mid + 1
print (f "{l} {r}" )
if __name__ = = "__main__" :
arr = [ 1 , 2 , 2 , 2 ]
K = 2
N = len (arr)
find_minrange(arr, N, K)
|
C#
using System;
class GFG {
static void find_minrange( int [] arr, int n, int k)
{
int [] count = new int [n + 1];
for ( int i = 0; i < arr.Length; i++)
count[arr[i]]++;
int [] prefixsum = new int [n + 1];
for ( int i = 1; i <= n; i++)
prefixsum[i] = prefixsum[i - 1] + count[i];
int low = 1, high = n;
int l = 0, r = 0;
while (low <= high) {
int mid = (low + high) / 2;
bool can = false ;
for ( int i = 1; i <= n - mid + 1; i++) {
int inrange = prefixsum[i + mid - 1]
- prefixsum[i - 1];
int outrange = n - inrange;
if (inrange - outrange >= k) {
can = true ;
l = i;
r = i + mid - 1;
break ;
}
}
if (can == true ) {
high = mid - 1;
}
else {
low = mid + 1;
}
}
Console.Write(l + " " + r);
}
public static void Main()
{
int [] arr = { 1, 2, 2, 2 };
int K = 2;
int N = arr.Length;
find_minrange(arr, N, K);
}
}
|
Javascript
<script>
function find_minrange(arr,n,k)
{
let count = new Array(n + 1).fill(0);
for (let x of arr)
count[x]++;
let prefixsum = new Array(n + 1).fill(0);
for (let i = 1; i <= n; i++)
prefixsum[i] = prefixsum[i - 1]
+ count[i];
let low = 1, high = n;
let l, r;
while (low <= high) {
let mid = Math.floor((low + high) / 2);
let can = false ;
for (let i = 1; i <= n - mid + 1; i++) {
let inrange
= prefixsum[i + mid - 1]
- prefixsum[i - 1];
let outrange = n - inrange;
if (inrange - outrange >= k) {
can = true ;
l = i;
r = i + mid - 1;
break ;
}
}
if (can) {
high = mid - 1;
}
else {
low = mid + 1;
}
}
document.write(l + " " + r);
}
let arr = [ 1, 2, 2, 2 ];
let K = 2;
let N = arr.length;
find_minrange(arr, N, K);
</script>
|
Time Complexity: O(N* log(N))
Auxiliary Space: O(N)
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