Strongest Fighter

Author: akashmodak97

Hacker Blocks/Strongest Fighter.cpp

@@ -0,0 +1,58 @@
+/* Hacker Blocks */
+/* Title - Strongest Fighter */
+/* Created By - Akash Modak */
+/* Date - 27/08/2020 */
+
+// There is a group of MMA fighters standing together in a line. Given the value of their strengths, find the strength of the strongest fighter in continuous sub-groups of size k.
+
+// Input Format
+// First line contains an integer N, the number of fighters Followed by N integers where i'th integer denotes the strength of i'th fighter. Next line contains the size of sub-group k
+
+// Constraints
+// 1<=N<=10^7
+// 1<=k<=N
+// 1 <= Ai <= 100000
+
+// Output Format
+// Space separated integers in a single line denoting strength of strongest fighters in the groups.
+
+// Sample Input
+// 5
+// 1 3 1 4 5
+// 3
+// Sample Output
+// 3 4 5
+// Explanation
+// First sub-group: 1 3 1 --> Max strength is 3
+// Second sub-group: 3 1 4 --> Max strength is 4
+// Third sub-group: 1 4 5 --> Max strength is 5
+
+#include<bits/stdc++.h>
+using namespace std;
+int main() {
+	int n,k;
+	cin>>n;
+	int a[n];
+	for(int i=0;i<n;i++)
+	cin>>a[i];
+	cin>>k;
+	int i;
+	deque<int> q;
+	for(i=0;i<k;i++)
+	{
+		while(!q.empty() and a[i]>a[q.back()])
+		q.pop_back();
+		q.push_back(i);
+	}
+	for(;i<n;i++){
+		cout<<a[q.front()]<<" ";
+		while(!q.empty() and q.front()<=i-k)
+		q.pop_front();
+
+		while(!q.empty() && a[i]>=a[q.back()])
+		q.pop_back();
+		q.push_back(i);
+	}
+	cout<<a[q.front()];
+	return 0;
+}