NUMBER OF PEOPLE VISIBLE IN A QUEUE | LEETCODE # 1944 | PYTHON MONOTONIC STACK SOLUTION

[@Q-ip2km]

Thank you! Very simple explanation

[@swethavarnaau7619]

Thanks for the most simple explanation. Could you please make a separate video on the monotonic stack algorithm .. ?

[@nishanksoni8964]

could u plz make video on sum of subarray ranges ? It's also a question of monotonic stack..

[@crackfaang]

Sure let me add it to my work queue. Make sure to subscribe so you don’t miss the video when it comes out

[@ram_luckyyy]

kzbin.info/door/eOjJmGArxh1pOpa3DBTWuA this is the you tube channel they explained very well

[@Nature-pb9dh]

Thank you!

[@joydeepsarkar4774]

if anybody is looking for a Java code...here it is int n = heights.length; int[] res = new int[n]; Stack st = new Stack(); st.push(heights[n-1]); res[n-1]=0; for(int i=n-2;i>=0;i--){ int visible = 0; while(st.size()>0 && heights[i]>st.peek()){ visible++; st.pop(); res[i] = visible; } if(st.size()>0){ visible++; res[i] = visible; } st.push(heights[i]); } return res;

[@sirfranciscanadianbacon1468]

// stores number of visible people per person in heights array int[] answer = new int[heights.length]; // last person cannot see anyone to the right answer[heights.length - 1] = 0; // keep track of heights of people visible to the right from current person Stack heightStack = new Stack(); // push last person's height onto stack, since iterating from right to left heightStack.push(heights[heights.length - 1]); // loop from right to left starting at second to last person for (int i = heights.length - 2; i >= 0; i--) { // reset visible count to zero int visiblePeople = 0; // while people still in stack and current person is taller than next while (heightStack.size() > 0 && heights[i] > heightStack.peek()) { // current person can see next person visiblePeople++; // remove shorter person from stack, since they will be blocked by current person heightStack.pop(); } // if stack not empty after loop ends (all shorter people have been removed), // there is at least one taller or equal-height person in stack who can be seen by current person if (heightStack.size() > 0) { // account for this person being visible visiblePeople++; } // store total number of people found visible for current person answer[i] = visiblePeople; // push current person's height onto stack for next comparison heightStack.push(heights[i]); } return answer;

[@subee128]

Thank you very much

[@ifuncuran]

"obviously we go from right to left" why?

[@Anonymous____________A721]

Because we want the next one, so we come from next to present, that is right to left