From 6a8019e2ca4ac5c5c1847cd4ed3df7d487894236 Mon Sep 17 00:00:00 2001
From: Hung Dang <hdang@athenahealth.com>
Date: Thu, 28 Mar 2019 19:56:33 -0400
Subject: [PATCH 1/3] Convert x - c >= 0 to x >= c since the former will be
 faster (https://godbolt.org/z/cB0bJm)

---
 chapter07.tex | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/chapter07.tex b/chapter07.tex
index 70fd873..6d11ac0 100644
--- a/chapter07.tex
+++ b/chapter07.tex
@@ -247,7 +247,7 @@ value[0] = 0;
 for (int x = 1; x <= n; x++) {
     value[x] = INF;
     for (auto c : coins) {
-        if (x-c >= 0) {
+        if (x >= c) {
             value[x] = min(value[x], value[x-c]+1);
         }
     }
@@ -283,7 +283,7 @@ value[0] = 0;
 for (int x = 1; x <= n; x++) {
     value[x] = INF;
     for (auto c : coins) {
-        if (x-c >= 0 && value[x-c]+1 < value[x]) {
+        if ((x >= c) && value[x-c]+1 < value[x]) {
             value[x] = value[x-c]+1;
             first[x] = c;
         }
@@ -358,7 +358,7 @@ for $0 \le x \le n$:
 count[0] = 1;
 for (int x = 1; x <= n; x++) {
     for (auto c : coins) {
-        if (x-c >= 0) {
+        if (x >= c) {
             count[x] += count[x-c];
         }
     }

From e4506fa0925341f061d8b93aa502cbb9b7b894c1 Mon Sep 17 00:00:00 2001
From: Hung Dang <hdang@athenahealth.com>
Date: Fri, 29 Mar 2019 00:56:46 -0400
Subject: [PATCH 2/3] Update the solution for the paths in a grid problem.

---
 chapter07.tex | 14 ++++++++++++--
 1 file changed, 12 insertions(+), 2 deletions(-)

diff --git a/chapter07.tex b/chapter07.tex
index 6d11ac0..30d3aad 100644
--- a/chapter07.tex
+++ b/chapter07.tex
@@ -634,12 +634,22 @@ int sum[N][N];
 \end{lstlisting}
 and calculate the sums as follows:
 \begin{lstlisting}
-for (int y = 1; y <= n; y++) {
-    for (int x = 1; x <= n; x++) {
+for (int y = 1; y < N; y++) {
+    for (int x = 1; x < N; x++) {
         sum[y][x] = max(sum[y][x-1],sum[y-1][x])+value[y][x];
     }
 }
 \end{lstlisting}
+However, below is a better solution that only uses one dimensional
+array to keep track of the maximum sums
+\begin{lstlisting}
+for (int y = 1; y < N; ++y) {
+    sum[0] += value[y][0];
+    for (int x = 1; x < N; ++x) {
+      sum[x] = value[y][x] + std::max(sum[x-1], sum[x]);
+    }
+}  
+\end{lstlisting}
 The time complexity of the algorithm is $O(n^2)$.
 
 \section{Knapsack problems}

From 2d53e3a6f5ea49d854c1525c537d1971e124d90a Mon Sep 17 00:00:00 2001
From: Hung Dang <hdang@athenahealth.com>
Date: Fri, 29 Mar 2019 00:58:55 -0400
Subject: [PATCH 3/3] Remove std namespace.

---
 chapter07.tex | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/chapter07.tex b/chapter07.tex
index 30d3aad..556d9be 100644
--- a/chapter07.tex
+++ b/chapter07.tex
@@ -646,7 +646,7 @@ array to keep track of the maximum sums
 for (int y = 1; y < N; ++y) {
     sum[0] += value[y][0];
     for (int x = 1; x < N; ++x) {
-      sum[x] = value[y][x] + std::max(sum[x-1], sum[x]);
+      sum[x] = value[y][x] + max(sum[x-1], sum[x]);
     }
 }  
 \end{lstlisting}