From 5f1ec0c5956b4747bc1410ef0b34fc24f15c2a48 Mon Sep 17 00:00:00 2001
From: Raku Zeta <zetaraku@gmail.com>
Date: Mon, 1 Jul 2024 17:45:59 +0800
Subject: [PATCH] Create 1985-find-the-kth-largest-integer-in-the-array.cpp

---
 ...d-the-kth-largest-integer-in-the-array.cpp | 41 +++++++++++++++++++
 1 file changed, 41 insertions(+)
 create mode 100644 code/medium/1985-find-the-kth-largest-integer-in-the-array.cpp

diff --git a/code/medium/1985-find-the-kth-largest-integer-in-the-array.cpp b/code/medium/1985-find-the-kth-largest-integer-in-the-array.cpp
new file mode 100644
index 0000000..24e07a5
--- /dev/null
+++ b/code/medium/1985-find-the-kth-largest-integer-in-the-array.cpp
@@ -0,0 +1,41 @@
+// Title: Find the Kth Largest Integer in the Array
+// Description:
+//     You are given an array of strings nums and an integer k.
+//     Each string in nums represents an integer without leading zeros.
+//     Return the string that represents the kth largest integer in nums.
+//     Note:
+//         Duplicate numbers should be counted distinctly.
+//         For example, if nums is ["1","2","2"], "2" is the first largest integer,
+//         "2" is the second-largest integer, and "1" is the third-largest integer.
+// Link: https://leetcode.com/problems/find-the-kth-largest-integer-in-the-array/
+
+// Time complexity: O(n)
+// Space complexity: O(1)
+class Solution {
+public:
+    std::string kthLargestNumber(std::vector<std::string> &nums, int k) {
+        auto target = nums.end() - k;
+
+        /*
+            void std::nth_element(first, nth, last, comp):
+                Rearranges the elements in the range [first,last),
+                in such a way that the element at the nth position is the element
+                that would be in that position in a sorted sequence.
+        */
+        std::nth_element(
+            nums.begin(), target, nums.end(),
+            [](std::string a, std::string b) {
+                // If the two numbers have different lengths,
+                // then the number with more digits is larger.
+                if (a.length() != b.length()) return a.length() < b.length();
+
+                // Otherwise, the two numbers have the same length,
+                // just compare them with lexicographical order.
+                // Note: The time complexity of this comparison is actually O(len(a)+len(b))
+                return a < b;
+            }
+        );
+
+        return *target;
+    }
+};