• Approach #1 Brute Force [Accepted]

Intuition

Do as directed in question

Algorithm

Hint: the number of ordered Numbers entered by the topic is strictly less than the second number
This problem is just based on the second number from the small to the big order, and then you start from the bottom table 0, and you go through the pairs,

Step 1: order the array pairs[][] according to the number of ordered pairs
Step 2: the variable i iterates from 0 to pairs.length-1,
Store the second number of the current ordered pairs in the temporary variable temp
Step 3: add the variable result to 1
Step 4: if the variable i is less than pairs.length and the number of the first number of ordered pairs is less than the current second number temp,
add the variable i to 1.Otherwise return step 2
Step 5: variable result is answer

Java

``````class Solution {
public int findLongestChain(int[][]pairs) {
int len=pairs.length;
int i=0;
int result=0;

Arrays.sort(pairs,(a,b)->(a[1]-b[1]));

while(i<len)
{
int temp=pairs[i][1];
result=result+1;
while(i<len&&pairs[i][0]<=temp)
i++;
}
return result;
}
}
``````

Complexity Analysis

Time complexity :O(n)

Space complexity :O(1) space required

• Hey the solution looks good, just that the time complexity will be O(nlogn) as you have sorting even though the while loop takes only O(n) to find the solution

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