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@@ -3702,14 +3702,14 @@
<p class="admonition-title">Question</p>
<p>Given an <span class="arithmatex">\(n \times m\)</span> two-dimensional grid <code>grid</code>, each cell in the grid contains a non-negative integer representing the cost of that cell. The robot starts from the top-left cell and can only move down or right at each step until it reaches the bottom-right cell. Return the minimum path sum from the top-left to the bottom-right.</p>
</div>
<p>The following figure shows an example, where the given grid's minimum path sum is <span class="arithmatex">\(13\)</span>.</p>
<p>Figure 14-10 shows an example, where the given grid's minimum path sum is <span class="arithmatex">\(13\)</span>.</p>
<p><a class="glightbox" href="../dp_solution_pipeline.assets/min_path_sum_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="Minimum Path Sum Example Data" class="animation-figure" src="../dp_solution_pipeline.assets/min_path_sum_example.png" /></a></p>
<p align="center"> Figure 14-10 &nbsp; Minimum Path Sum Example Data </p>
<p><strong>First step: Think about each round of decisions, define the state, and thereby obtain the <span class="arithmatex">\(dp\)</span> table</strong></p>
<p>Each round of decisions in this problem is to move one step down or right from the current cell. Suppose the row and column indices of the current cell are <span class="arithmatex">\([i, j]\)</span>, then after moving down or right, the indices become <span class="arithmatex">\([i+1, j]\)</span> or <span class="arithmatex">\([i, j+1]\)</span>. Therefore, the state should include two variables: the row index and the column index, denoted as <span class="arithmatex">\([i, j]\)</span>.</p>
<p>The state <span class="arithmatex">\([i, j]\)</span> corresponds to the subproblem: the minimum path sum from the starting point <span class="arithmatex">\([0, 0]\)</span> to <span class="arithmatex">\([i, j]\)</span>, denoted as <span class="arithmatex">\(dp[i, j]\)</span>.</p>
<p>Thus, we obtain the two-dimensional <span class="arithmatex">\(dp\)</span> matrix shown below, whose size is the same as the input grid <span class="arithmatex">\(grid\)</span>.</p>
<p>Thus, we obtain the two-dimensional <span class="arithmatex">\(dp\)</span> matrix shown in Figure 14-11, whose size is the same as the input grid <span class="arithmatex">\(grid\)</span>.</p>
<p><a class="glightbox" href="../dp_solution_pipeline.assets/min_path_sum_solution_state_definition.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="State definition and DP table" class="animation-figure" src="../dp_solution_pipeline.assets/min_path_sum_solution_state_definition.png" /></a></p>
<p align="center"> Figure 14-11 &nbsp; State definition and DP table </p>
@@ -3720,7 +3720,7 @@
</div>
<p><strong>Second step: Identify the optimal substructure, then derive the state transition equation</strong></p>
<p>For the state <span class="arithmatex">\([i, j]\)</span>, it can only be derived from the cell above <span class="arithmatex">\([i-1, j]\)</span> or the cell to the left <span class="arithmatex">\([i, j-1]\)</span>. Therefore, the optimal substructure is: the minimum path sum to reach <span class="arithmatex">\([i, j]\)</span> is determined by the smaller of the minimum path sums of <span class="arithmatex">\([i, j-1]\)</span> and <span class="arithmatex">\([i-1, j]\)</span>.</p>
<p>Based on the above analysis, the state transition equation shown in the following figure can be derived:</p>
<p>Based on the above analysis, the state transition equation shown in Figure 14-12 can be derived:</p>
<div class="arithmatex">\[
dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
\]</div>
@@ -3734,7 +3734,7 @@ dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
</div>
<p><strong>Third step: Determine boundary conditions and state transition order</strong></p>
<p>In this problem, the states in the first row can only come from the states to their left, and the states in the first column can only come from the states above them, so the first row <span class="arithmatex">\(i = 0\)</span> and the first column <span class="arithmatex">\(j = 0\)</span> are the boundary conditions.</p>
<p>As shown in the Figure 14-13 , since each cell is derived from the cell to its left and the cell above it, we use loops to traverse the matrix, the outer loop iterating over the rows and the inner loop iterating over the columns.</p>
<p>As shown in Figure 14-13, since each cell is derived from the cell to its left and the cell above it, we use loops to traverse the matrix, the outer loop iterating over the rows and the inner loop iterating over the columns.</p>
<p><a class="glightbox" href="../dp_solution_pipeline.assets/min_path_sum_solution_initial_state.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="Boundary conditions and state transition order" class="animation-figure" src="../dp_solution_pipeline.assets/min_path_sum_solution_initial_state.png" /></a></p>
<p align="center"> Figure 14-13 &nbsp; Boundary conditions and state transition order </p>
@@ -4015,7 +4015,7 @@ dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
<p><div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dfs%28grid%3A%20list%5Blist%5Bint%5D%5D,%20i%3A%20int,%20j%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E6%9A%B4%E5%8A%9B%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E4%B8%BA%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%8D%95%E5%85%83%E6%A0%BC%EF%BC%8C%E5%88%99%E7%BB%88%E6%AD%A2%E6%90%9C%E7%B4%A2%0A%20%20%20%20if%20i%20%3D%3D%200%20and%20j%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E8%8B%A5%E8%A1%8C%E5%88%97%E7%B4%A2%E5%BC%95%E8%B6%8A%E7%95%8C%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20%2B%E2%88%9E%20%E4%BB%A3%E4%BB%B7%0A%20%20%20%20if%20i%20%3C%200%20or%20j%20%3C%200%3A%0A%20%20%20%20%20%20%20%20return%20inf%0A%20%20%20%20%23%20%E8%AE%A1%E7%AE%97%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i-1,%20j%29%20%E5%92%8C%20%28i,%20j-1%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20up%20%3D%20min_path_sum_dfs%28grid,%20i%20-%201,%20j%29%0A%20%20%20%20left%20%3D%20min_path_sum_dfs%28grid,%20i,%20j%20-%201%29%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i,%20j%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20return%20min%28left,%20up%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%20%23%20%E6%9A%B4%E5%8A%9B%E6%90%9C%E7%B4%A2%0A%20%20%20%20res%20%3D%20min_path_sum_dfs%28grid,%20n%20-%201,%20m%20-%201%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dfs%28grid%3A%20list%5Blist%5Bint%5D%5D,%20i%3A%20int,%20j%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E6%9A%B4%E5%8A%9B%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E4%B8%BA%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%8D%95%E5%85%83%E6%A0%BC%EF%BC%8C%E5%88%99%E7%BB%88%E6%AD%A2%E6%90%9C%E7%B4%A2%0A%20%20%20%20if%20i%20%3D%3D%200%20and%20j%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E8%8B%A5%E8%A1%8C%E5%88%97%E7%B4%A2%E5%BC%95%E8%B6%8A%E7%95%8C%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20%2B%E2%88%9E%20%E4%BB%A3%E4%BB%B7%0A%20%20%20%20if%20i%20%3C%200%20or%20j%20%3C%200%3A%0A%20%20%20%20%20%20%20%20return%20inf%0A%20%20%20%20%23%20%E8%AE%A1%E7%AE%97%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i-1,%20j%29%20%E5%92%8C%20%28i,%20j-1%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20up%20%3D%20min_path_sum_dfs%28grid,%20i%20-%201,%20j%29%0A%20%20%20%20left%20%3D%20min_path_sum_dfs%28grid,%20i,%20j%20-%201%29%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i,%20j%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20return%20min%28left,%20up%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%20%23%20%E6%9A%B4%E5%8A%9B%E6%90%9C%E7%B4%A2%0A%20%20%20%20res%20%3D%20min_path_sum_dfs%28grid,%20n%20-%201,%20m%20-%201%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen &gt;</a></div></p>
</details>
<p>The following figure shows the recursive tree rooted at <span class="arithmatex">\(dp[2, 1]\)</span>, which includes some overlapping subproblems, the number of which increases sharply as the size of the grid <code>grid</code> increases.</p>
<p>Figure 14-14 shows the recursive tree rooted at <span class="arithmatex">\(dp[2, 1]\)</span>, which includes some overlapping subproblems, the number of which increases sharply as the size of the grid <code>grid</code> increases.</p>
<p>Essentially, the reason for overlapping subproblems is: <strong>there are multiple paths to reach a certain cell from the top-left corner</strong>.</p>
<p><a class="glightbox" href="../dp_solution_pipeline.assets/min_path_sum_dfs.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="Brute-force search recursive tree" class="animation-figure" src="../dp_solution_pipeline.assets/min_path_sum_dfs.png" /></a></p>
<p align="center"> Figure 14-14 &nbsp; Brute-force search recursive tree </p>
@@ -4358,7 +4358,7 @@ dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
<p><div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dfs_mem%28%0A%20%20%20%20grid%3A%20list%5Blist%5Bint%5D%5D,%20mem%3A%20list%5Blist%5Bint%5D%5D,%20i%3A%20int,%20j%3A%20int%0A%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E4%B8%BA%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%8D%95%E5%85%83%E6%A0%BC%EF%BC%8C%E5%88%99%E7%BB%88%E6%AD%A2%E6%90%9C%E7%B4%A2%0A%20%20%20%20if%20i%20%3D%3D%200%20and%20j%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E8%8B%A5%E8%A1%8C%E5%88%97%E7%B4%A2%E5%BC%95%E8%B6%8A%E7%95%8C%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20%2B%E2%88%9E%20%E4%BB%A3%E4%BB%B7%0A%20%20%20%20if%20i%20%3C%200%20or%20j%20%3C%200%3A%0A%20%20%20%20%20%20%20%20return%20inf%0A%20%20%20%20%23%20%E8%8B%A5%E5%B7%B2%E6%9C%89%E8%AE%B0%E5%BD%95%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%0A%20%20%20%20if%20mem%5Bi%5D%5Bj%5D%20!%3D%20-1%3A%0A%20%20%20%20%20%20%20%20return%20mem%5Bi%5D%5Bj%5D%0A%20%20%20%20%23%20%E5%B7%A6%E8%BE%B9%E5%92%8C%E4%B8%8A%E8%BE%B9%E5%8D%95%E5%85%83%E6%A0%BC%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20up%20%3D%20min_path_sum_dfs_mem%28grid,%20mem,%20i%20-%201,%20j%29%0A%20%20%20%20left%20%3D%20min_path_sum_dfs_mem%28grid,%20mem,%20i,%20j%20-%201%29%0A%20%20%20%20%23%20%E8%AE%B0%E5%BD%95%E5%B9%B6%E8%BF%94%E5%9B%9E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i,%20j%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20mem%5Bi%5D%5Bj%5D%20%3D%20min%28left,%20up%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%20%20%20%20return%20mem%5Bi%5D%5Bj%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%23%20%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%0A%20%20%20%20mem%20%3D%20%5B%5B-1%5D%20*%20m%20for%20_%20in%20range%28n%29%5D%0A%20%20%20%20res%20%3D%20min_path_sum_dfs_mem%28grid,%20mem,%20n%20-%201,%20m%20-%201%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=16&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
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</details>
<p>As shown in the Figure 14-15 , after introducing memoization, all subproblem solutions only need to be calculated once, so the time complexity depends on the total number of states, i.e., the grid size <span class="arithmatex">\(O(nm)\)</span>.</p>
<p>As shown in Figure 14-15, after introducing memoization, all subproblem solutions only need to be calculated once, so the time complexity depends on the total number of states, i.e., the grid size <span class="arithmatex">\(O(nm)\)</span>.</p>
<p><a class="glightbox" href="../dp_solution_pipeline.assets/min_path_sum_dfs_mem.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="Memoized search recursive tree" class="animation-figure" src="../dp_solution_pipeline.assets/min_path_sum_dfs_mem.png" /></a></p>
<p align="center"> Figure 14-15 &nbsp; Memoized search recursive tree </p>
@@ -4716,7 +4716,7 @@ dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
<p><div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dp%28grid%3A%20list%5Blist%5Bint%5D%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20m%20for%20_%20in%20range%28n%29%5D%0A%20%20%20%20dp%5B0%5D%5B0%5D%20%3D%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%0A%20%20%20%20for%20j%20in%20range%281,%20m%29%3A%0A%20%20%20%20%20%20%20%20dp%5B0%5D%5Bj%5D%20%3D%20dp%5B0%5D%5Bj%20-%201%5D%20%2B%20grid%5B0%5D%5Bj%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B0%5D%20%3D%20dp%5Bi%20-%201%5D%5B0%5D%20%2B%20grid%5Bi%5D%5B0%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%E5%92%8C%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20m%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20min%28dp%5Bi%5D%5Bj%20-%201%5D,%20dp%5Bi%20-%201%5D%5Bj%5D%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%20%20%20%20return%20dp%5Bn%20-%201%5D%5Bm%20-%201%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20min_path_sum_dp%28grid%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
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</details>
<p>The following figures show the state transition process of the minimum path sum, traversing the entire grid, <strong>thus the time complexity is <span class="arithmatex">\(O(nm)\)</span></strong>.</p>
<p>Figure 14-16 show the state transition process of the minimum path sum, traversing the entire grid, <strong>thus the time complexity is <span class="arithmatex">\(O(nm)\)</span></strong>.</p>
<p>The array <code>dp</code> is of size <span class="arithmatex">\(n \times m\)</span>, <strong>therefore the space complexity is <span class="arithmatex">\(O(nm)\)</span></strong>.</p>
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