Sync LeetCode submission Runtime - 293 ms (47.37%), Memory - 48.3 MB (82.81%)

Author: jimit105

1995-finding-pairs-with-a-certain-sum/README.md

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+<p>You are given two integer arrays <code>nums1</code> and <code>nums2</code>. You are tasked to implement a data structure that supports queries of two types:</p>
+
+<ol>
+	<li><strong>Add</strong> a positive integer to an element of a given index in the array <code>nums2</code>.</li>
+	<li><strong>Count</strong> the number of pairs <code>(i, j)</code> such that <code>nums1[i] + nums2[j]</code> equals a given value (<code>0 &lt;= i &lt; nums1.length</code> and <code>0 &lt;= j &lt; nums2.length</code>).</li>
+</ol>
+
+<p>Implement the <code>FindSumPairs</code> class:</p>
+
+<ul>
+	<li><code>FindSumPairs(int[] nums1, int[] nums2)</code> Initializes the <code>FindSumPairs</code> object with two integer arrays <code>nums1</code> and <code>nums2</code>.</li>
+	<li><code>void add(int index, int val)</code> Adds <code>val</code> to <code>nums2[index]</code>, i.e., apply <code>nums2[index] += val</code>.</li>
+	<li><code>int count(int tot)</code> Returns the number of pairs <code>(i, j)</code> such that <code>nums1[i] + nums2[j] == tot</code>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input</strong>
+[&quot;FindSumPairs&quot;, &quot;count&quot;, &quot;add&quot;, &quot;count&quot;, &quot;count&quot;, &quot;add&quot;, &quot;add&quot;, &quot;count&quot;]
+[[[1, 1, 2, 2, 2, 3], [1, 4, 5, 2, 5, 4]], [7], [3, 2], [8], [4], [0, 1], [1, 1], [7]]
+<strong>Output</strong>
+[null, 8, null, 2, 1, null, null, 11]
+
+<strong>Explanation</strong>
+FindSumPairs findSumPairs = new FindSumPairs([1, 1, 2, 2, 2, 3], [1, 4, 5, 2, 5, 4]);
+findSumPairs.count(7);  // return 8; pairs (2,2), (3,2), (4,2), (2,4), (3,4), (4,4) make 2 + 5 and pairs (5,1), (5,5) make 3 + 4
+findSumPairs.add(3, 2); // now nums2 = [1,4,5,<strong><u>4</u></strong><code>,5,4</code>]
+findSumPairs.count(8);  // return 2; pairs (5,2), (5,4) make 3 + 5
+findSumPairs.count(4);  // return 1; pair (5,0) makes 3 + 1
+findSumPairs.add(0, 1); // now nums2 = [<strong><u><code>2</code></u></strong>,4,5,4<code>,5,4</code>]
+findSumPairs.add(1, 1); // now nums2 = [<code>2</code>,<strong><u>5</u></strong>,5,4<code>,5,4</code>]
+findSumPairs.count(7);  // return 11; pairs (2,1), (2,2), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,4) make 2 + 5 and pairs (5,3), (5,5) make 3 + 4
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums1.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= nums2.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums1[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>1 &lt;= nums2[i] &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= index &lt; nums2.length</code></li>
+	<li><code>1 &lt;= val &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= tot &lt;= 10<sup>9</sup></code></li>
+	<li>At most <code>1000</code> calls are made to <code>add</code> and <code>count</code> <strong>each</strong>.</li>
+</ul>

1995-finding-pairs-with-a-certain-sum/solution.py

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+# Approach: Hash Table
+
+from collections import Counter
+
+class FindSumPairs:
+
+    def __init__(self, nums1: List[int], nums2: List[int]):
+        self.nums1 = nums1
+        self.nums2 = nums2
+        self.cnt = Counter(nums2)
+        
+
+    def add(self, index: int, val: int) -> None:
+        _nums2, _cnt = self.nums2, self.cnt
+        _cnt[_nums2[index]] -= 1
+        _nums2[index] += val
+        _cnt[_nums2[index]] += 1
+        
+
+    def count(self, tot: int) -> int:
+        _nums1, _cnt = self.nums1, self.cnt
+        ans = 0
+        for num in _nums1:
+            if (rest := tot - num) in _cnt:
+                ans += _cnt[rest]
+        return ans
+
+
+# Your FindSumPairs object will be instantiated and called as such:
+# obj = FindSumPairs(nums1, nums2)
+# obj.add(index,val)
+# param_2 = obj.count(tot)