Runtime: 62 ms (Top 92.08%) | Memory: 43.30 MB (Top 49.95%)

Author: AnasImloul

scripts/algorithms/S/Sum of Subarray Minimums/Sum of Subarray Minimums.cpp

@@ -1,68 +1,24 @@
-class Solution {
-public: //Thanks to Bhalerao-2002
-    int sumSubarrayMins(vector<int>& A) {
-        
-        int n = A.size();
-        int MOD = 1e9 + 7;
-        vector<int> left(n), right(n);
-        
-        // for every i find the Next smaller element to left and right
-        
-        // Left
-        stack<int>st;
-        st.push(0);
-        left[0] = 1; // distance = 1, left not found, this is distance multiplied with num, so it can't be zero
-        for(int i=1; i<n; i++)
+// Runtime: 62 ms (Top 92.08%) | Memory: 43.30 MB (Top 49.95%)
+
+class Solution 
+{
+public:
+    int sumSubarrayMins(vector<int>& n) 
+    {
+        vector<long> s, sums(n.size(),0);
+        long j, res=0, mod = 1000000007;
+        for (int i = 0; i < n.size(); ++i)
         {
-            while(!st.empty() && A[i] < A[st.top()]) 
-                st.pop();
-            
-            if(st.empty()) 
-                left[i] = i+1; // total distance if less element not found = i+1
-            else 
-                left[i] = i - st.top(); // distance = i-st.top()
-            
-            st.push(i);
+            while (!s.empty() && n[s.back()] > n[i])
+                s.pop_back();
+            j = !s.empty() ? s.back() : -1;
 
+            sums[i] = ((j>=0?sums[j]:0) + (i-j)*n[i]) % mod;
+            s.push_back(i);
         }
-        
-        while(st.size()) 
-            st.pop();
-        
-        // Right 
-        st.push(n-1);
-        right[n-1] = 1; // distance = 1, right not found, this is distance multiplied with num, so it can't be zero
-        for(int i=n-2; i>=0; i--)
-        {
-            while(!st.empty() && A[i] <= A[st.top()]) 
-                st.pop();
-            
-            if(st.empty()) 
-                right[i] = n-i; // distance
-            else 
-                right[i] = st.top()-i;
-            
-            st.push(i);
-        }
-        
-        // total number of subarrays : (Left[i] * Right[i])
-        // total contribution in A[i] element in final answer : (Left * Right) * A[i] 
-        
-        for(int i=0; i<n; i++) 
-             cout << left[i] << " : " << right[i] << endl;
-        
-        // for each i, contribution is (Left * Right) * Element 
-        
-        int res = 0;
-        for(int i=0; i<n; i++)
-        {
-            long long prod = (left[i]*right[i])%MOD;
-            prod = (prod*A[i])%MOD;
-            res = (res + prod)%MOD;
-        }
-        
-        return res%MOD;
-    }
-    
 
-};
+        for (int i = 0; i < sums.size(); ++i)
+            res = (res + sums[i]) % mod;
+        return res;
+    }
+};