Improved task 794.

Author: javadev

src/main/java/g0701_0800/s0794_valid_tic_tac_toe_state/Solution.java

@@ -1,66 +1,63 @@
 package g0701_0800.s0794_valid_tic_tac_toe_state;
 
-// #Medium #Array #String #2022_03_26_Time_0_ms_(100.00%)_Space_42.1_MB_(13.47%)
+// #Medium #Array #String #2022_03_30_Time_0_ms_(100.00%)_Space_39.6_MB_(91.75%)
 
 public class Solution {
     public boolean validTicTacToe(String[] board) {
-        int cO = 0;
-        int cX = 0;
-        for (int x = 0; x < 3; x++) {
-            for (int y = 0; y < 3; y++) {
-                char ch = board[x].charAt(y);
-                if (ch == 'O') {
-                    cO++;
-                } else if (ch == 'X') {
-                    cX++;
+        // X=1,O=-1,’ ’=0
+        int sum = 0;
+        int[] winsCol = new int[3];
+        int[] winsDiag = new int[2];
+        boolean xWin = false;
+        boolean oWin = false;
+        for (int i = 0; i < 3; i++) {
+            String str = board[i];
+            int rowSum = 0;
+            for (int j = 0; j < 3; j++) {
+                // char chr=str.toCharArray()[j];
+                int intchr = 0;
+                if (str.toCharArray()[j] == 'X') {
+                    intchr = 1;
                 }
+                if (str.toCharArray()[j] == 'O') {
+                    intchr = -1;
+                }
+                rowSum += intchr;
+                winsCol[j] += intchr;
+                if (i == 2 && winsCol[j] == 3) {
+                    xWin = true;
+                }
+                if (i == 2 && winsCol[j] == -3) {
+                    oWin = true;
+                }
+                if (Math.abs(i - j) != 1) {
+                    if (i == j && i == 1) {
+                        winsDiag[0] += intchr;
+                        winsDiag[1] += intchr;
+                    } else if (i == j) {
+                        winsDiag[0] += intchr;
+                    } else {
+                        winsDiag[1] += intchr;
+                    }
+                }
+                if (i == 2 && Math.max(winsDiag[0], winsDiag[1]) == 3) {
+                    xWin = true;
+                }
+                if (i == 2 && Math.min(winsDiag[0], winsDiag[1]) == -3) {
+                    oWin = true;
+                }
+            }
+            if (rowSum == 3) {
+                xWin = true;
             }
+            if (rowSum == -3) {
+                oWin = true;
+            }
+            sum += rowSum;
         }
-        char a1 = board[0].charAt(0);
-        char a2 = board[0].charAt(1);
-        char a3 = board[0].charAt(2);
-        char a4 = board[1].charAt(0);
-        char a5 = board[1].charAt(1);
-        char a6 = board[1].charAt(2);
-        char a7 = board[2].charAt(0);
-        char a8 = board[2].charAt(1);
-        char a9 = board[2].charAt(2);
-        // The main thing here is we just need to check if the given sequence is possible or not.
-        // We dont need to find any winner. *Logic is the one who has last chance can be in 1 row,
-        // column(horizontally,vertically or diagonally) not the other one.
-        // So few cases will arise
-        // 1) Count of 'X' is 1 more than count of 'O'-> Only X can win and 'O' cannot win (Last
-        // Chance is with X so only X can win)
-        // 2) Count of 'X' is equal to count of 'O'   -> Only 'O' can win and 'X' cannot win(Check
-        // few cases)(Last Chance is with 'O' so only 'O' can win)
-        // 3) return false (rest all are invalid case)
-
-        // Eg - if X=4 and O=3 (ie Count of 'O' is 1 less) then 'O' cannot win and
-        // if we find such a case then we need to return false else the sequence is fine(try
-        // yourself & see)
-        if ((cX - 1) == cO) {
-            // These are all horizaontal, vertical, diagonal
-            return (a1 != a2 || a2 != a3 || a1 != 'O')
-                    && (a4 != a5 || a5 != a6 || a4 != 'O')
-                    && (a7 != a8 || a8 != a9 || a7 != 'O')
-                    && (a1 != a5 || a5 != a9 || a1 != 'O')
-                    && (a3 != a5 || a5 != a7 || a3 != 'O')
-                    && (a1 != a4 || a4 != a7 || a1 != 'O')
-                    && (a2 != a5 || a5 != a8 || a2 != 'O')
-                    && (a3 != a6 || a6 != a9 || a3 != 'O');
-        } else if (cX == cO) {
-            // if X=4 and O=4 (ie both r equal) then 'X' cannot win amd
-            // if we find such a case then we need to return false else the sequence is fine(try
-            // yourself & see)
-            return (a1 != a2 || a2 != a3 || a1 != 'X')
-                    && (a4 != a5 || a5 != a6 || a4 != 'X')
-                    && (a7 != a8 || a8 != a9 || a7 != 'X')
-                    && (a1 != a5 || a5 != a9 || a1 != 'X')
-                    && (a3 != a5 || a5 != a7 || a3 != 'X')
-                    && (a1 != a4 || a4 != a7 || a1 != 'X')
-                    && (a2 != a5 || a5 != a8 || a2 != 'X')
-                    && (a3 != a6 || a6 != a9 || a3 != 'X');
+        if (sum == 0 && !xWin) {
+            return true;
         }
-        return false;
+        return sum == 1 && !oWin;
     }
 }