Given an array arr[] of positive integers of size N and a positive integer K, the task is to find the maximum possible length of a subarray which can be made equal by adding some integer value to each element of the sub-array such that the sum of the added elements does not exceed K.
Examples:
Input: N = 5, arr[] = {1, 4, 9, 3, 6}, K = 9
Output: 3
Explanation:
{1, 4} : {1+3, 4} = {4, 4}
{4, 9} : {4+5, 9} = {9, 9}
{3, 6} : {3+3, 6} = {6, 6}
{9, 3, 6} : {9, 3+6, 6+3} = {9, 9, 9}
Hence, the maximum length of such a subarray is 3.Input: N = 6, arr[] = {2, 4, 7, 3, 8, 5}, K = 10
Output: 4
Approach: This problem can be solved by using dynamic programming.
- Initialize:
- dp[]: Stores the sum of elements that are added to the subarray.
- deque: Stores the indices of the maximum element for each subarray.
- pos: Index of the current position of the subarray.
- ans: Length of the maximum subarray.
- mx: Maximum element of a subarray
- pre: Previous index of the current subarray.
- Traverse the array and check if the deque is empty or not. If yes, then update the maximum element and the index of the maximum element along with the indices of pre and pos.
- Check if the currently added element is greater than K. If yes, then remove it from dp[] array and update the indices of pos and pre.
- Finally, update the maximum length of the valid sub-array.
Below is the implementation of the above approach:
// C++ code for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find maximum // possible length of subarray int validSubArrLength( int arr[],
int N, int K)
{ // Stores the sum of elements
// that needs to be added to
// the sub array
int dp[N + 1];
// Stores the index of the
// current position of subarray
int pos = 0;
// Stores the maximum
// length of subarray.
int ans = 0;
// Maximum element from
// each subarray length
int mx = 0;
// Previous index of the
// current subarray of
// maximum length
int pre = 0;
// Deque to store the indices
// of maximum element of
// each sub array
deque< int > q;
// For each array element,
// find the maximum length of
// required subarray
for ( int i = 0; i < N; i++) {
// Traverse the deque and
// update the index of
// maximum element.
while (!q.empty()
&& arr[q.back()] < arr[i])
q.pop_back();
q.push_back(i);
// If it is first element
// then update maximum
// and dp[]
if (i == 0) {
mx = arr[i];
dp[i] = arr[i];
}
// Else check if current
// element exceeds max
else if (mx <= arr[i]) {
// Update max and dp[]
dp[i] = dp[i - 1] + arr[i];
mx = arr[i];
}
else {
dp[i] = dp[i - 1] + arr[i];
}
// Update the index of the
// current maximum length
// subarray
if (pre == 0)
pos = 0;
else
pos = pre - 1;
// While current element
// being added to dp[] array
// exceeds K
while ((i - pre + 1) * mx
- (dp[i] - dp[pos])
> K
&& pre < i) {
// Update index of
// current position and
// the previous position
pos = pre;
pre++;
// Remove elements
// from deque and
// update the
// maximum element
while (!q.empty()
&& q.front() < pre
&& pre < i) {
q.pop_front();
mx = arr[q.front()];
}
}
// Update the maximum length
// of the required subarray.
ans = max(ans, i - pre + 1);
}
return ans;
} // Driver Program int main()
{ int N = 6;
int K = 8;
int arr[] = { 2, 7, 1, 3, 4, 5 };
cout << validSubArrLength(arr, N, K);
return 0;
} |
// Java code for the above approach import java.util.*;
class GFG{
// Function to find maximum // possible length of subarray static int validSubArrLength( int arr[],
int N, int K)
{ // Stores the sum of elements
// that needs to be added to
// the sub array
int []dp = new int [N + 1 ];
// Stores the index of the
// current position of subarray
int pos = 0 ;
// Stores the maximum
// length of subarray.
int ans = 0 ;
// Maximum element from
// each subarray length
int mx = 0 ;
// Previous index of the
// current subarray of
// maximum length
int pre = 0 ;
// Deque to store the indices
// of maximum element of
// each sub array
Deque<Integer> q = new LinkedList<>();
// For each array element,
// find the maximum length of
// required subarray
for ( int i = 0 ; i < N; i++)
{
// Traverse the deque and
// update the index of
// maximum element.
while (!q.isEmpty() &&
arr[q.getLast()] < arr[i])
q.removeLast();
q.add(i);
// If it is first element
// then update maximum
// and dp[]
if (i == 0 )
{
mx = arr[i];
dp[i] = arr[i];
}
// Else check if current
// element exceeds max
else if (mx <= arr[i])
{
// Update max and dp[]
dp[i] = dp[i - 1 ] + arr[i];
mx = arr[i];
}
else
{
dp[i] = dp[i - 1 ] + arr[i];
}
// Update the index of the
// current maximum length
// subarray
if (pre == 0 )
pos = 0 ;
else
pos = pre - 1 ;
// While current element
// being added to dp[] array
// exceeds K
while ((i - pre + 1 ) * mx -
(dp[i] - dp[pos]) > K && pre < i)
{
// Update index of
// current position and
// the previous position
pos = pre;
pre++;
// Remove elements from
// deque and update the
// maximum element
while (!q.isEmpty() &&
q.peek() < pre && pre < i)
{
q.removeFirst();
mx = arr[q.peek()];
}
}
// Update the maximum length
// of the required subarray.
ans = Math.max(ans, i - pre + 1 );
}
return ans;
} // Driver code public static void main(String[] args)
{ int N = 6 ;
int K = 8 ;
int arr[] = { 2 , 7 , 1 , 3 , 4 , 5 };
System.out.print(validSubArrLength(arr, N, K));
} } // This code is contributed by amal kumar choubey |
# Python3 code for the above approach # Function to find maximum # possible length of subarray def validSubArrLength(arr, N, K):
# Stores the sum of elements
# that needs to be added to
# the sub array
dp = [ 0 for i in range (N + 1 )]
# Stores the index of the
# current position of subarray
pos = 0
# Stores the maximum
# length of subarray.
ans = 0
# Maximum element from
# each subarray length
mx = 0
# Previous index of the
# current subarray of
# maximum length
pre = 0
# Deque to store the indices
# of maximum element of
# each sub array
q = []
# For each array element,
# find the maximum length of
# required subarray
for i in range (N):
# Traverse the deque and
# update the index of
# maximum element.
while ( len (q) and arr[ len (q) - 1 ] < arr[i]):
q.remove(q[ len (q) - 1 ])
q.append(i)
# If it is first element
# then update maximum
# and dp[]
if (i = = 0 ):
mx = arr[i]
dp[i] = arr[i]
# Else check if current
# element exceeds max
elif (mx < = arr[i]):
# Update max and dp[]
dp[i] = dp[i - 1 ] + arr[i]
mx = arr[i]
else :
dp[i] = dp[i - 1 ] + arr[i]
# Update the index of the
# current maximum length
# subarray
if (pre = = 0 ):
pos = 0
else :
pos = pre - 1
# While current element
# being added to dp[] array
# exceeds K
while ((i - pre + 1 ) *
mx - (dp[i] - dp[pos]) > K and
pre < i):
# Update index of
# current position and
# the previous position
pos = pre
pre + = 1
# Remove elements
# from deque and
# update the
# maximum element
while ( len (q) and
q[ 0 ] < pre and
pre < i):
q.remove(q[ 0 ])
mx = arr[q[ 0 ]]
# Update the maximum length
# of the required subarray.
ans = max (ans, i - pre + 1 )
return ans
# Driver code if __name__ = = '__main__' :
N = 6
K = 8
arr = [ 2 , 7 , 1 , 3 , 4 , 5 ]
print (validSubArrLength(arr, N, K))
# This code is contributed by ipg2016107 |
<script> // Javascript code for the above approach // Function to find maximum // possible length of subarray function validSubArrLength(arr, N, K)
{ // Stores the sum of elements
// that needs to be added to
// the sub array
var dp = Array(N+1);
// Stores the index of the
// current position of subarray
var pos = 0;
// Stores the maximum
// length of subarray.
var ans = 0;
// Maximum element from
// each subarray length
var mx = 0;
// Previous index of the
// current subarray of
// maximum length
var pre = 0;
// Deque to store the indices
// of maximum element of
// each sub array
var q = [];
// For each array element,
// find the maximum length of
// required subarray
for ( var i = 0; i < N; i++) {
// Traverse the deque and
// update the index of
// maximum element.
while (q.length!=0
&& arr[q[q.length-1]] < arr[i])
q.pop();
q.push(i);
// If it is first element
// then update maximum
// and dp[]
if (i == 0) {
mx = arr[i];
dp[i] = arr[i];
}
// Else check if current
// element exceeds max
else if (mx <= arr[i]) {
// Update max and dp[]
dp[i] = dp[i - 1] + arr[i];
mx = arr[i];
}
else {
dp[i] = dp[i - 1] + arr[i];
}
// Update the index of the
// current maximum length
// subarray
if (pre == 0)
pos = 0;
else
pos = pre - 1;
// While current element
// being added to dp[] array
// exceeds K
while ((i - pre + 1) * mx
- (dp[i] - dp[pos])
> K
&& pre < i) {
// Update index of
// current position and
// the previous position
pos = pre;
pre++;
// Remove elements
// from deque and
// update the
// maximum element
while (q.length!=0
&& q[0] < pre
&& pre < i) {
q.shift();
mx = arr[q[0]];
}
}
// Update the maximum length
// of the required subarray.
ans = Math.max(ans, i - pre + 1);
}
return ans;
} // Driver Program var N = 6;
var K = 8;
var arr = [2, 7, 1, 3, 4, 5];
document.write( validSubArrLength(arr, N, K)); </script> |
// C# program to implement above approach using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{ // Function to find maximum
// possible length of subarray
static int validSubArrLength( int [] arr, int N, int K)
{
// Stores the sum of elements
// that needs to be added to
// the sub array
int [] dp = new int [N + 1];
// Stores the index of the
// current position of subarray
int pos = 0;
// Stores the maximum
// length of subarray.
int ans = 0;
// Maximum element from
// each subarray length
int mx = 0;
// Previous index of the
// current subarray of
// maximum length
int pre = 0;
// Deque to store the indices
// of maximum element of
// each sub array
List< int > q = new List< int >();
// For each array element,
// find the maximum length of
// required subarray
for ( int i = 0 ; i < N ; i++)
{
// Traverse the deque and
// update the index of
// maximum element.
while (q.Count > 0 && arr[q[q.Count - 1]] < arr[i]){
q.RemoveAt(q.Count - 1);
}
q.Add(i);
// If it is first element
// then update maximum
// and dp[]
if (i == 0)
{
mx = arr[i];
dp[i] = arr[i];
}
// Else check if current
// element exceeds max
else if (mx <= arr[i])
{
// Update max and dp[]
dp[i] = dp[i - 1] + arr[i];
mx = arr[i];
}
else
{
dp[i] = dp[i - 1] + arr[i];
}
// Update the index of the
// current maximum length
// subarray
if (pre == 0)
pos = 0;
else
pos = pre - 1;
// While current element
// being added to dp[] array
// exceeds K
while ((i - pre + 1) * mx - (dp[i] - dp[pos]) > K && pre < i)
{
// Update index of
// current position and
// the previous position
pos = pre;
pre++;
// Remove elements from
// deque and update the
// maximum element
while (q.Count > 0 && q[0] < pre && pre < i)
{
q.RemoveAt(0);
mx = arr[q[0]];
}
}
// Update the maximum length
// of the required subarray.
ans = Math.Max(ans, i - pre + 1);
}
return ans;
}
// Driver code
public static void Main( string [] args){
int N = 6;
int K = 8;
int [] arr = new int []{ 2, 7, 1, 3, 4, 5 };
Console.WriteLine(validSubArrLength(arr, N, K));
}
} |
4
Time Complexity: O(N2)
Auxiliary Space Complexity: O(N)