From 66e4183a9fc3437f20d20ddf83f2aba495ea3f31 Mon Sep 17 00:00:00 2001
From: sudiptarathi2020 <rathisudipta99@gmail.com>
Date: Fri, 17 Jan 2025 01:02:48 +0600
Subject: [PATCH] Added English tutorial for 1093 #512

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+
+## [LightOj-1093](https://lightoj.com/problem/ghajini)
+## Problem Summary
+Amir is playing a game called "Find Max Difference". But he can only remember numbers for a short period of time, that is `d` milliseconds.
+The game involves:
+
+- A sequence of integers appearing one at a time, for 1 millisecond each.
+- The goal is to calculate the maximum difference between any two integers that Amir can remember(i.e., within the last `d` milliseconds).
+- Input Constrains
+	*  T <= 5 (Number of test case)
+	*  2 <= n <=10^5 (Total number of integers the screen shows)
+	*  1 <= d <= n (Amir memory duration)
+	*  Sequence values: 0 <= a[i] <= 10^8
+
+
+
+
+
+## Solution
+
+- We will iterate the array. In each index we compare the minimum and maximum numbers from the last `d` integers. Amir remembers to maximize the difference and stores the max difference in a variable.
+- A naive approach of iterating through the last `d` numbers for every index would be too slow((O(n * d)).
+- We can efficiently find the minimum and maximum of the last `d` numbers using sparse table or segment tree. Then the complexity would be O(logn).
+
+### C++ code using sparse table
+
+```c++
+#include<bits/stdc++.h>
+using namespace std;
+const int maxN = 1e5+1;
+const int maxLog = 20;
+int stMax[maxN][maxLog];
+int stMin[maxN][maxLog];
+int lg[maxN];
+void init(int arr[],int n){
+    for(int i=0;i<n;i++){
+        stMax[i][0] = arr[i];
+        stMin[i][0] = arr[i];
+    }
+    for(int j=1;j<maxLog;j++){
+        for(int i=0;i+(1<<j)<=n;i++){
+            stMax[i][j] = max(stMax[i][j-1],stMax[i+(1<<(j-1))][j-1]);
+            stMin[i][j] = min(stMin[i][j-1],stMin[i+(1<<(j-1))][j-1]);
+        }
+    }
+}
+int queryMax(int l,int r){
+    if(l<0)l=0;
+    int j = lg[r-l+1];
+    return max(stMax[l][j],stMax[r-(1<<j)+1][j]);
+}
+int queryMin(int l,int r){
+    if(l<0)l=0;
+    int j = lg[r-l+1];
+    return min(stMin[l][j],stMin[r-(1<<j)+1][j]);
+}
+int main(){
+    ios::sync_with_stdio(false);
+    cin.tie(0);cout.tie(0);
+    lg[1] = 0;
+    for(int i=2;i<maxN;i++){
+        lg[i] = lg[i/2]+1;
+    }
+    int tcs;
+    cin>>tcs;
+    for(int tc=1;tc<=tcs;tc++){
+        int n,d;
+        cin>>n>>d;
+        int arr[n];
+        for(int i=0;i<n;i++){
+            cin>>arr[i];
+        }
+        init(arr,n);
+        int ans = 0;
+        for(int i=0;i<n;i++){
+            ans = max(ans, queryMax(i-d+1,i)-queryMin(i-d+1,i));
+        }
+        cout<<"Case "<<tc<<": "<<ans<<endl;
+    }
+}
+```
+
+## Resources
+- [Sparse Table cp-algorithms](https://cp-algorithms.com/data_structures/sparse-table.html)
+- [geeksforgeeks](https://www.geeksforgeeks.org/sparse-table/)
+
+## Author
+[Sudipta Singha](https://github.com/sudiptarathi2020)