From a72c7df7acb3a4f029865ffaf3cbb266f736f13d Mon Sep 17 00:00:00 2001
From: Delta09 <Delta09@users.noreply.github.com>
Date: Tue, 26 Mar 2019 16:35:45 -0700
Subject: [PATCH] Add Second Solution

A second solution with O(NM) time complexity and O(1) space complexity
---
 src/chapter1/ch1-q8.js | 56 +++++++++++++++++++++++++++++++++++++++++-
 1 file changed, 55 insertions(+), 1 deletion(-)

diff --git a/src/chapter1/ch1-q8.js b/src/chapter1/ch1-q8.js
index cbdbc5d..7acabf8 100644
--- a/src/chapter1/ch1-q8.js
+++ b/src/chapter1/ch1-q8.js
@@ -1,6 +1,6 @@
 'use strict';
 
-/**
+/**  1st Solution
  * Do a first pass through the matrix to find which cells have 0's. When a 0 is
  * found then mark that row and column as needing to be zeroed out. On the second
  * pass zero out any cells that need to be zeroed out based on the row or column
@@ -47,3 +47,57 @@ export function zeroMatrix(matrix) {
 
   return matrix;
 }
+
+
+/** 2nd Solution
+ * Do a first pass through the matrix to find which cells have 0's. When a 0 is 
+ * found then mark it in the first column and row. Then check the first row to  
+ * see which columns to zero out. Repeat the process for the first column to 
+ * see which rows to zero out.
+ *
+ * N = matrix Y dimension
+ * M = matrix X dimension
+ * Time: O(N * M)
+ * Additional space: O(1)
+ *
+ * @param  {array} matrix Matrix to be zeroed in-place
+ * @return {array}        Matrix that has been zeroed, same object as input
+ */
+
+const zero = (matrix) => {
+
+    let rows = matrix.length,
+        cols = matrix[0].length;
+    
+    //First pass through to indicate which cols and row to zero out
+    for (let i = 0; i < rows; ++i) {
+        for (let j = 0; j < cols; ++j) {
+            if (matrix[i][j] === 0) {
+                matrix[i][0] = 0
+                matrix[0][j] = 0;
+            }
+        }
+    }
+  
+    //zero out colums by looking at first row
+    for (let j = 0; j < cols; ++j) {
+        if (matrix[0][j] == 0) {
+            for (let i = 0; i < rows; ++i) {
+                matrix[i][j] = 0;
+            }
+        }
+    }
+
+    //zero out rows by looking at first col
+    for (let i = 0; i < cols; ++i) {
+        if (matrix[i][0] == 0) {
+            for (let j = 0; j < rows; ++j) {
+                matrix[i][j] = 0;
+            }
+        }
+    }
+
+    return matrix;
+}
+
+