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Minimum_Length_Subarray_Sum_atleast_K.java
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104 lines (93 loc) · 2.2 KB
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package matrix;
import java.util.*;
public class Minimum_Length_Subarray_Sum_atleast_K{
public static Scanner ss=new Scanner(System.in);
public static Random rr=new Random();
//print array
public static void print(int arr[])
{
for(int i=0;i<arr.length-1;i++)
{
System.out.print(arr[i]+",");
}
System.out.print(arr[arr.length-1]);
System.out.println();
}
/*
In the Sample Input 1, there are many subarrays with a sum greater than 9.
But the subarrays with sum equal to 9 are {5, 3, 1}, {2, 4, 3}, and {3, 6}.
As the subarray {3, 6} has minimum length; the output is 2.
*/
/*
Minimum Length Subarray with Sum at least K
Given an array A of N integers, and an integer K,
write a program to find the length of minimum length subarray whose sum is greater
than or equal to K
*/
public static int fun(int arr[])
{
System.out.println("Enter the minimum sum");
int key=ss.nextInt();
int count=0;
int min_len=arr.length+1;
for(int i=0;i<arr.length;i++)
{
int cur=arr[i];
if(cur>=key)
{
count++;
return 1;
}
for(int j=i+1;j<arr.length;j++)
{
cur=cur+arr[j];
if(cur>=key && (j-i+1)<min_len)
{
min_len=(j-i+1);
count++;
}
}
}
if(count==0)
{
return 0;
}
else{
return min_len;
}
}
//time complexity of O(n)
public static int fun(int arr[],int key)
{
int ans=Integer.MAX_VALUE;
int l=0;
int sum=0;
for(int i=0;i<arr.length;i++)
{
sum=sum+arr[i];
while(sum>=key)
{
ans=Math.min(ans,i+1-l);
sum=sum-arr[l];
l++;
}
}
if(ans!=Integer.MAX_VALUE)
{
return ans;
}
else{
return 0;
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int n=ss.nextInt();
int arr[]=new int[n];
for(int i=0;i<arr.length;i++)
{
arr[i]=rr.nextInt(100);
}
print(arr);
}
}