Software Engineer

How to use lps[] to decide the next positions (or to know the number of characters to be skipped)?

We start the comparison of pat[j] with j = 0 with characters of the current window of text.
We keep matching characters txt[i] and pat[j] and keep incrementing i and j while pat[j] and txt[i] keep matching.
When we see a mismatch
We know that characters pat[0..j-1] match with txt[i-j…i-1] (Note that j starts with 0 and increments it only when there is a match).
We also know (from the above definition) that lps[j-1] is the count of characters of pat[0…j-1] that are both proper prefix and suffix.
From the above two points, we can conclude that we do not need to match these lps[j-1] characters with txt[i-j…i-1] because we know that these characters will anyway match. Let us consider the above example to understand this.

// C++ implementation of simple method to find count of
// pairs with given sum K.
#include 
using namespace std;

// Returns number of pairs in arr[0..n-1] with sum equal
// to 'K'
int getPairsCount(int arr[], int n, int K)
{
// Initialize result
int count = 0;

// Consider all possible pairs and check their sums
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (arr[i] + arr[j] == K)
count++;

return count;
}

// Driver function to test the above function
int main()
{
int arr[] = { 1, 5, 7, -1 };
int n = sizeof(arr) / sizeof(arr[0]);
int K = 6;
cout << "Count of pairs is "
// Function Call
<< getPairsCount(arr, n, K);
return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

// C++ implementation to find the index of
// first '1' in a sorted array of 0's and 1's
#include 
using namespace std;

// function to find the index of first '1'
int indexOfFirstOne(int arr[], int n)
{
// traverse the array from left to right
for (int i = 0; i < n; i++)

// if true, then return i
if (arr[i] == 1)
return i;

// 1's are not present in the array
return -1;
}

// Driver program to test above
int main()
{
int arr[] = { 0, 0, 0, 0, 0, 0, 1, 1, 1, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << indexOfFirstOne(arr, n);
return 0;
}

class Solution {
public:
bool repeatedSubstringPattern(string s) {
int n=s.size();
for(int i=n/2;i>=1;i--){ 
if(n%i==0){ 
if(s.substr(0,n-i)==s.substr(i))return true; 
}
}
return false;
}
};

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