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LeetCode/1753-path-with-minimum-effort/README.md

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+<p>You are a hiker preparing for an upcoming hike. You are given <code>heights</code>, a 2D array of size <code>rows x columns</code>, where <code>heights[row][col]</code> represents the height of cell <code>(row, col)</code>. You are situated in the top-left cell, <code>(0, 0)</code>, and you hope to travel to the bottom-right cell, <code>(rows-1, columns-1)</code> (i.e.,&nbsp;<strong>0-indexed</strong>). You can move <strong>up</strong>, <strong>down</strong>, <strong>left</strong>, or <strong>right</strong>, and you wish to find a route that requires the minimum <strong>effort</strong>.</p>
+
+<p>A route&#39;s <strong>effort</strong> is the <strong>maximum absolute difference</strong><strong> </strong>in heights between two consecutive cells of the route.</p>
+
+<p>Return <em>the minimum <strong>effort</strong> required to travel from the top-left cell to the bottom-right cell.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2020/10/04/ex1.png" style="width: 300px; height: 300px;" /></p>
+
+<pre>
+<strong>Input:</strong> heights = [[1,2,2],[3,8,2],[5,3,5]]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.
+This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2020/10/04/ex2.png" style="width: 300px; height: 300px;" /></p>
+
+<pre>
+<strong>Input:</strong> heights = [[1,2,3],[3,8,4],[5,3,5]]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/10/04/ex3.png" style="width: 300px; height: 300px;" />
+<pre>
+<strong>Input:</strong> heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> This route does not require any effort.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>rows == heights.length</code></li>
+	<li><code>columns == heights[i].length</code></li>
+	<li><code>1 &lt;= rows, columns &lt;= 100</code></li>
+	<li><code>1 &lt;= heights[i][j] &lt;= 10<sup>6</sup></code></li>
+</ul>

LeetCode/1753-path-with-minimum-effort/solution.cpp

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+class Solution {
+private:
+    int dx[4]={-1,0,1,0};
+    int dy[4]={0,1,0,-1};
+public:
+    int minimumEffortPath(vector<vector<int>>& heights) {
+
+        int low=0;
+        int high=1e6;
+        int ans=0;
+        while(low<=high)
+        {
+            int mid=(low+high)/2;
+
+            if(check(heights,mid))
+            {
+                ans=mid;
+                high=mid-1;
+            }
+            else
+            {
+                low=mid+1;
+            }
+        }
+        return ans;
+
+    }
+
+    bool check(vector<vector<int>>&heights,int mid)
+    {
+        int r=heights.size();
+        int c=heights[0].size();
+        int i,j;
+
+        vector<vector<int>>vis(r,vector<int>(c,0));
+        vis[0][0]=1;
+        queue<pair<int,int>>q;
+        q.push({0,0});
+        while(!q.empty())
+        {
+            int x=q.front().first;
+            int y=q.front().second;
+            q.pop();
+            if(x==r-1 and y==c-1)
+                return true;
+            for(int k=0;k<4;k++)
+            {
+                int nx=dx[k]+x;
+                int ny=dy[k]+y;
+                if(nx>=0 and ny>=0 and nx<r and ny<c and !vis[nx][ny])
+                {
+                    int newDistance=abs(heights[nx][ny]-heights[x][y]);
+
+                    if(newDistance<=mid)
+                    {
+                        q.push({nx,ny});
+                        vis[nx][ny]=1;
+                    }
+
+                }
+            }
+        }
+        return false;
+    }
+};