Add property based on longestCommonSubstring

Author: dubzzz

src/algorithms/string/longest-common-substring/__test__/longestCommonSubstring.test.js

@@ -1,3 +1,4 @@
+import * as fc from 'fast-check';
 import longestCommonSubstring from '../longestCommonSubstring';
 
 describe('longestCommonSubstring', () => {
@@ -8,4 +9,73 @@ describe('longestCommonSubstring', () => {
     expect(longestCommonSubstring('ABABC', 'BABCA')).toBe('BABC');
     expect(longestCommonSubstring('BABCA', 'ABCBA')).toBe('ABC');
   });
+  it('should find same length whatever the ordering of parameters [property]', () => {
+    fc.assert(
+      fc.property(
+        fc.fullUnicodeString(),
+        fc.fullUnicodeString(),
+        (s1, s2) => longestCommonSubstring(s1, s2).length === longestCommonSubstring(s2, s1).length
+      )
+    );
+  });
+  it('should be a substring of both strings [property]', () => {
+    fc.assert(
+      fc.property(
+        fc.fullUnicodeString(),
+        fc.fullUnicodeString(),
+        (s1, s2) => {
+          const longest = longestCommonSubstring(s1, s2);
+          return s1.includes(longest) && s2.includes(longest);
+        }
+      )
+    );
+  });
+  it('should find the substring when its the whole string [property]', () => {
+    fc.assert(
+      fc.property(
+        fc.fullUnicodeString(),
+        fc.fullUnicodeString(),
+        fc.fullUnicodeString(),
+        (s, prefix, suffix) => longestCommonSubstring(prefix + s + suffix, s) === s
+      )
+    );
+  });
+  it('should find a substring greater or equal to the one we know [property]', () => {
+    fc.assert(
+      fc.property(
+        fc.fullUnicodeString(),
+        fc.fullUnicodeString(),
+        fc.fullUnicodeString(),
+        fc.fullUnicodeString(),
+        fc.fullUnicodeString(),
+        (s, prefix1, suffix1, prefix2, suffix2) => {
+          const s1 = prefix1 + s + suffix1;
+          const s2 = prefix2 + s + suffix2;
+          const match = longestCommonSubstring(s1, s2);
+          return [...s].length <= [...match].length;
+        }
+      )
+    );
+  });
+});
+it('should find the longest common when two distinct areas match [property]', () => {
+  fc.assert(
+    fc.property(
+      fc.fullUnicodeString(1, 10),
+      fc.fullUnicodeString(1, 10),
+      fc.fullUnicodeString(),
+      fc.fullUnicodeString(),
+      fc.fullUnicodeString(),
+      fc.fullUnicodeString(),
+      fc.fullUnicodeString(),
+      fc.fullUnicodeString(),
+      fc.fullUnicodeString(),
+      (c1, c2, prefix1, mid1, suffix1, prefix2, mid2, suffix2) => {
+        const s1 = prefix1 + c1 + mid1 + c2 + suffix1;
+        const s2 = prefix2 + c1 + mid2 + c2 + suffix2;
+        const match = longestCommonSubstring(s1, s2);
+        return Math.max([...c1].length, [...c2].length) <= [...match].length;
+      }
+    )
+  );
 });