Create Z-Algorithm

String_Algo/Z-Algorithm

@@ -0,0 +1,80 @@
+#include <bits/stdc++.h>
+using namespace std;
+#define ll long long
+#define ln "\n"
+#define pb push_back
+#define pll pair<ll,ll>
+#define ppll pair<ll,pll>
+#define vll vector<ll>
+#define vs vector<string>
+#define vpll vector<pll>
+#define vvll vector<vector<ll>>
+#define vvpll vector<vpll>
+#define f first
+#define s second
+#define bs binary_search
+#define lb lower_bound
+#define ub upper_bound
+#define Test ll T;cin>>T; while(T--)
+#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL);
+#define init(x,n,v) for(ll i=0;i<=n;i++) x[i]=v;
+#define all(x) x.begin(),x.end()
+#define allr(x) x.rbegin(),x.rend()
+#define pi 3.14159265358979323846
+#define MOD 998244353
+#define MAX 2000000000000000000
+#define MAXN 100005
+
+vector<ll>  zfunction(string &s){
+
+    // 0th -index based z-function
+    //  IN z-function at each index we check the longest common prefix between that index and starting point 
+    // we take the z-function of the starting index as 0 
+    // Here we maintain a segment l to r which are similar to the 0th index to r-l index
+    // when we are in that segment we don't have to traverse to r because the same answer for corresponding starting element is already known
+    // we just have to traverse when the common prefix is greater than or equal to the range (r-index) or we are outside the right boundary r
+   // Time complexity for this algorithm is O(n) and to search a pattern in this we just add that pattern before the main text/array in that case the size become n+m(size of pattern) and complexity goes to O(n+m)
+
+    ll n = s.size(); 
+    vector<ll> zarr(n,0);
+    for(ll i=1,l=0,r=0;i<n;i++){
+
+        if(i>r){  // we are outside the segment [ l - r ]
+           l=r=i; // we have to start traversing from here and compare the elements to corresponding starting element till we get any mismatch or end of the text/array 
+           while(r<n && s[r-l]==s[r])r++;
+           --r;  // since the breaking point is mismatched so we have to decrement the r by 1
+           zarr[i] = r-l+1;
+        }
+        else {  // we are inside the segment [ l - r ]
+           ll k = i - l;
+           if(zarr[k]<r-i+1) zarr[i] = zarr[i-l]; // if length of common prefix is less than (r-i+1) 
+           else{ // if common prefix is equal to or greater than (r-i+1)
+            l=i;
+            while(r<n && s[r-l]==s[r])r++;
+            --r;  // since the breaking point is mismatched so we have to decrement the r by 1
+            zarr[i] = r-l+1;
+           }
+        }
+    }
+    // we just return the Z-array 
+    return zarr;
+}
+
+int main() {
+
+   string s ="ababcbababca";
+
+   //  string s for which we are going to make z array 
+
+   vector<ll> zarr = zfunction(s);  // calling Z-function
+
+   // printing z array
+
+   for(ll i=0;i<s.size();i++){
+      cout<<zarr[i]<<"  ";
+   }    
+
+//   OUTPUT : 0  0  2  0  0  0  5  0  2  0  0  1 
+
+   return 0;
+}