Refine 668_Kth_Smallest_Number_in_Multiplication_Table

Author: qiyuangong

README.md

@@ -175,6 +175,7 @@ I'm currently working on [Analytics-Zoo](https://github.com/intel-analytics/anal
 | 628 | [Maximum Product of Three Numbers](https://leetcode.com/problems/maximum-product-of-three-numbers/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/628_Maximum_Product_of_Three_Numbers.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/628_Maximum_Product_of_Three_Numbers.java) | Actually, we should only care about min1, min2 and max1-max3, to find these five elements, we can use 1. Brute force, O(n^3) and O(1)<br>2. Sort, O(nlogn) and O(1)<br>3. Single scan with 5 elements keep, O(n) and O(1) |
 | 654 | [Maximum Binary Tree](https://leetcode.com/problems/maximum-binary-tree/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/654_Maximum_Binary_Tree.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/654_Maximum_Binary_Tree.java) | 1. Divide and conquer, recursive, O(n^2)<br>2. Monotonic stack, O(n) |
 | 665 | [Non-decreasing Array](https://leetcode.com/problems/non-decreasing-array/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/665_Non-decreasing_Array.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/665_Non-decreasing_Array.java) | 1. Find the broken index, then check this point, O(n) and O(1)<br>2. Replace broken point with correct value, then check if there are more than 1 broken point, O(n) and O(1) |
+| 668 | [Kth Smallest Number in Multiplication Table](https://leetcode.com/problems/kth-smallest-number-in-multiplication-table/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/668_Kth_Smallest_Number_in_Multiplication_Table.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/668_Kth_Smallest_Number_in_Multiplication_Table.java) [Cpp](https://github.com/qiyuangong/leetcode/blob/master/cpp/668_Kth_Smallest_Number_in_Multiplication_Table.cpp) | Binary search, O(mlog(mn) and O(1) |
 | 671 | [Second Minimum Node In a Binary Tree](https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/671_Second_Minimum_Node_In_a_Binary_Tree.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/671_Second_Minimum_Node_In_a_Binary_Tree.java) | Note that min value is root: 1. Get all values then find result, O(n) and O(n)<br>2. BFS or DFS traverse the tree, then find the reslut, O(n) and O(n)|
 | 674 | [Longest Continuous Increasing Subsequence](https://leetcode.com/problems/longest-continuous-increasing-subsequence/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/674_Longest_Continuous_Increasing_Subsequence.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/674_Longest_Continuous_Increasing_Subsequence.java) | Scan nums once, check nums[i] < nums[i+1], if not reset count, O(n) and O(1) |
 | 680 | [Valid Palindrome II](https://leetcode.com/problems/valid-palindrome-ii/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/680_Valid_Palindrome_II.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/680_Valid_Palindrome_II.java) | Recursively check s[left == end, when not equal delete left or right. |

cpp/668_Kth_Smallest_Number_in_Multiplication_Table.cpp

@@ -1,26 +1,27 @@
 class Solution {
 public:
-    #define ll long long int
-    
-    bool valid(ll x, int m, int n, int k){
-        int cnt=0;
-        for(int i=1;i<=m;i++){
-            cnt+=n<x/i?n:x/i;
-            if(x/i==0)break;
+#define ll long long int
+    bool valid(ll x, int m, int n, int k) {
+        int cnt = 0;
+        for (int i = 1; i <= m; i++) {
+            cnt += n < x / i ? n : x / i;
+            if (x / i == 0)
+                break;
         }
-        return cnt>=k;
+        return cnt >= k;
     }
-    
+
     int findKthNumber(int n1, int n2, int k) {
-        ll l=0, r=n1*n2,ans;
-        while(l<=r){
-            ll m = l +(r-l)/2;
-            if(valid(m,n1,n2,k)){
-                ans=m;
-                r=m-1;
-            }
-            else{
-                l=m+1;
+        ll l = 0, r = n1 * n2, ans;
+        while (l <= r) {
+            // ith row [i, 2*i, 3*i, ..., n*i]
+            // for each column, k = x // i
+            ll m = l + (r - l) / 2;
+            if (valid(m, n1, n2, k)) {
+                ans = m;
+                r = m - 1;
+            } else {
+                l = m + 1;
             }
         }
         return ans;

java/668_Kth_Smallest_Number_in_Multiplication_Table.java

@@ -0,0 +1,22 @@
+class Solution {
+    // https://leetcode.com/problems/kth-smallest-number-in-multiplication-table/solution/
+    public boolean enough(int x, int m, int n, int k) {
+        int count = 0;
+        for (int i = 1; i <= m; i++) {
+            count += Math.min(x / i, n);
+        }
+        return count >= k;
+    }
+
+    public int findKthNumber(int m, int n, int k) {
+        int lo = 1, hi = m * n;
+        while (lo < hi) {
+            // ith row [i, 2*i, 3*i, ..., n*i]
+            // for each column, k = x // i
+            int mi = lo + (hi - lo) / 2;
+            if (!enough(mi, m, n, k)) lo = mi + 1;
+            else hi = mi;
+        }
+        return lo;
+    }
+}

python/668_Kth_Smallest_Number_in_Multiplication_Table.py

@@ -0,0 +1,19 @@
+class Solution:
+    def findKthNumber(self, m: int, n: int, k: int) -> int:
+        # https://leetcode.com/problems/kth-smallest-number-in-multiplication-table/solution/
+        def enough(x):
+            count = 0
+            # ith row [i, 2*i, 3*i, ..., n*i]
+            # for each column, k = x // i
+            for i in range(1, m+1):
+                count += min(x // i, n)
+            return count >= k
+
+        lo, hi = 1, m * n
+        while lo < hi:
+            mi = (lo + hi) // 2
+            if not enough(mi):
+                lo = mi + 1
+            else:
+                hi = mi
+        return lo