Create 2040.Kth-Smallest-Product-of-Two-Sorted-Arrays_v2.cpp

rajanwalia4tech · Oct 17, 2021 · c5aeb89 · c5aeb89
1 parent 25c3e9f
commit c5aeb89
Showing 1 changed file with 75 additions and 0 deletions.
diff --git a/...allest-Product-of-Two-Sorted-Arrays/2040.Kth-Smallest-Product-of-Two-Sorted-Arrays_v2.cpp b/...allest-Product-of-Two-Sorted-Arrays/2040.Kth-Smallest-Product-of-Two-Sorted-Arrays_v2.cpp
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+using LL = long long;
+class Solution {
+public:
+    long long kthSmallestProduct(vector<int>& nums1, vector<int>& nums2, long long k) 
+    {
+        if (nums1.size() > nums2.size())
+            return kthSmallestProduct(nums2, nums1, k);
+
+        LL left = -1e10, right = 1e10;
+        while (left < right)
+        {
+            LL mid = left+(right-left)/2;
+            LL count = countSmallerOrEqual(mid, nums1, nums2);
+            if (count < k)
+                left = mid+1;
+            else
+                right = mid;                
+        }
+        return left;
+    }
+
+    LL countSmallerOrEqual(LL mid, vector<int>& nums1, vector<int>& nums2)
+    {
+        LL ret = 0;
+
+        if (mid >= 0)
+        {
+            int j0 = nums2.size()-1;
+            int j1 = nums2.size()-1;
+
+            for (int i=0; i<nums1.size(); i++)
+            {
+                if (nums1[i]>0)
+                {
+                    while (j0>=0 && (LL)nums1[i]*(LL)nums2[j0] > mid)
+                        j0--;
+                    ret += j0+1;
+                }
+                else if (nums1[i]==0)
+                    ret += nums2.size();
+                else
+                {
+                    while (j1>=0 && (LL)nums1[i]*(LL)nums2[j1] <= mid)
+                        j1--;
+                    ret += nums2.size()-1-j1;
+                }
+            }
+        }
+        else
+        {
+            int j0 = 0;
+            int j1 = 0;
+
+            for (int i=0; i<nums1.size(); i++)
+            {
+                if (nums1[i]>0)
+                {
+                    while (j0<nums2.size() && (LL)nums1[i]*(LL)nums2[j0] <= mid)
+                        j0++;
+                    ret += j0;
+                }
+                else if (nums1[i]==0)
+                    ret += 0;
+                else
+                {
+                    while (j1<nums2.size() && (LL)nums1[i]*(LL)nums2[j1] > mid)
+                        j1++;
+                    ret += nums2.size()-j1;
+                }
+            }
+        }
+
+        return ret;
+    }
+};