Image Overlap

Author: Cappuccinuo

Array/835. Image Overlap/README.md

@@ -0,0 +1,26 @@
+Time spent : 5 min(See hint)
+
+GIVEN: Two images A and B, represented as binary, square matrices of the same size
+
+RETURNS: the largest possible overlap
+
+EXAMPLES:
+
+**Example 1:**
+
+```
+Input: A = [[1,1,0],
+            [0,1,0],
+            [0,1,0]]
+       B = [[0,0,0],
+            [0,1,1],
+            [0,0,1]]
+Output: 3
+Explanation: We slide A to right by 1 unit and down by 1 unit.
+```
+
+Algorithm:
+
+We can see from the example that in A and B, A\[0][0] and B\[1][1] can be at the same position, A\[0][0] and B\[1][2]  can be at the same position, so what we can count is the delta between x and y of same 1 in A and B.   So we need to record the delta, if we say delta is 3, then their are 3 unit that are overlap.
+
+e.g. In the example, A\[0][0] and B\[1][1], A\[0][1] and B\[1][2], A\[1][1] and B\[2][2].

Array/835. Image Overlap/Solution.java

@@ -0,0 +1,28 @@
+class Solution {
+    public int largestOverlap(int[][] A, int[][] B) {
+        int N = A.length;
+        int[][] count = new int[2 * N + 1][2 * N + 1];
+        
+        for (int i = 0; i < N; i++) {
+            for (int j = 0; j < N; j++) {
+                if (A[i][j] == 1) {
+                    for (int k = 0; k < N; k++) {
+                        for (int l = 0; l < N; l++) {
+                            if (B[k][l] == 1) {
+                                count[k - i + N][l - j + N] += 1;
+                            } 
+                        }
+                    }
+                }
+            }
+        }
+        
+        int ans = 0;
+        for (int[] val : count) {
+            for (int y : val) {
+                ans = Math.max(ans, y);
+            }
+        }
+        return ans;
+    }
+}

HashTable/205. Isomorphic Strings/Solution.java

@@ -14,4 +14,32 @@ public boolean isIsomorphic(String s, String t) {
         }
         return true;
     }
+}
+
+class Solution {
+    public boolean isIsomorphic(String s, String t) {
+        if (s.length() != t.length()) {
+            return false;
+        }
+        int len = s.length();
+        Map<Character, Character> sToT = new HashMap<>();
+        Map<Character, Character> tToS = new HashMap<>();
+        for (int i = 0; i < len; i++) {
+            char a = s.charAt(i);
+            char b = t.charAt(i);
+            if (!sToT.containsKey(a) && !tToS.containsKey(b)) {
+                sToT.put(a, b);
+                tToS.put(b, a);
+            }
+            else {
+                if (!sToT.containsKey(a) || !tToS.containsKey(b)) {
+                    return false;
+                }
+                else if (sToT.get(a) != b || tToS.get(b) != a) {
+                    return false;
+                }
+            }
+        }
+        return true;
+    }
 }