From 67f7191089a3560023cb8ecc110ba20427d54df2 Mon Sep 17 00:00:00 2001
From: Karan <karandeepbatra@gmail.com>
Date: Tue, 16 Oct 2018 10:16:50 +0530
Subject: [PATCH] Add Z-Algo

---
 String/Z-Algo/z_algo.cpp | 62 ++++++++++++++++++++++++++++++++++++++++
 1 file changed, 62 insertions(+)
 create mode 100644 String/Z-Algo/z_algo.cpp

diff --git a/String/Z-Algo/z_algo.cpp b/String/Z-Algo/z_algo.cpp
new file mode 100644
index 00000000..601e7184
--- /dev/null
+++ b/String/Z-Algo/z_algo.cpp
@@ -0,0 +1,62 @@
+#include <bits/stdc++.h>
+using namespace std;
+typedef long long ll;
+ 
+#define TRACE
+#ifdef TRACE
+#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
+template <typename Arg1>
+void __f(const char* name, Arg1&& arg1){
+  cerr << name << " : " << arg1 << std::endl;
+}
+template <typename Arg1, typename... Args>
+void __f(const char* names, Arg1&& arg1, Args&&... args){
+  const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
+}
+#else
+#define trace(...)
+#endif
+
+#define MOD                 1000000007LL
+#define EPS                 1e-9
+#define io                  ios_base::sync_with_stdio(false);cin.tie(NULL);
+
+const int N = 1e5 + 5;
+
+// z[i] = the length of the longest substring of 's' starting at index 'i' which is at the same time a prefix of the string 's'
+int z[N];
+
+int n;
+
+void z_func(string &s){			// l, r are the invariant variables => [l,r] is the maximum interval such that s[l,r] is a prefix substring
+    int l = 0, r = 0;
+    n = s.size();
+    for(int i = 1; i < n; i++){
+        if(i > r){				// make a new interval l,r and find the substring starting at i which is a prefix substring
+            l = r = i;
+            while(r < n && s[r - l] == s[r])    r++;
+            r--;
+            z[i] = r - l + 1;
+        }else{					// try to expand the current interval by observing the value of z[k]
+            int k = i - l;
+            if(z[k] < r - i + 1)    z[i] = z[k];		// can't be more than z[k] since s[i..r] matches with s[k...z[k]]
+            else{										// we can expand the current r since z[k] allows us to do so
+                l = i;
+                while(r < n && s[r - l] == s[r])    r++;
+                r--;
+                z[i] = r - l + 1;
+            }
+        }
+    }
+    return;
+}
+
+
+int main(){
+	io;
+	string s = "aaaaa";
+	z_func(s);
+	for(int i = 1; i < s.size(); i++)
+		trace(i, z[i]);
+	return 0;
+}
\ No newline at end of file