Peak element

Arrays-algos/mountainSubarrayProblem.cpp

@@ -0,0 +1,71 @@
+// https://practice.geeksforgeeks.org/problems/mountain-subarray-problem/1
+
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution{
+  public:
+    vector<bool> processQueries(int a[], int n, vector<pair<int, int>> &queries,
+                                int q) {
+
+        vector<bool>ans;
+        int left[n], right[n];
+        left[0]=0;
+        right[n-1]=n-1;
+        int k=0;
+
+        // left[i] stores the last index on left side which is increasing 
+        // i.e. greater than its previous element
+        for(int i=1; i<n; i++)
+            {
+                if(a[i-1]<a[i])
+                    k=i;
+                left[i]=k;
+            }
+        // right[i] will store the first index on the right side which is decreasing 
+        // i.e. greater than its next element. 
+        for(int i=n-2; i>=0; i--)  
+        {
+            if(a[i+1]<a[i])
+                k=i;
+            right[i]=k;
+
+        }
+
+        for(auto x: queries)
+        {
+            int l= x.first;
+            int r= x.second;
+
+            if(left[r]<=right[l])
+                ans.push_back(true);
+            else
+                ans.push_back(false);
+        }
+    return ans;
+    }
+};
+
+int main() {
+    int tc;
+    cin >> tc;
+    while (tc--) {
+        int n, i, q;
+        cin >> n;
+        int a[n];
+        for (i = 0; i < n; i++) {
+            cin >> a[i];
+        }
+        cin >> q;
+        vector<pair<int, int>> queries(q);
+        for (i = 0; i < q; i++) {
+            cin >> queries[i].first >> queries[i].second;
+        }
+        Solution obj;
+        auto v = obj.processQueries(a, n, queries, q);
+        for (bool u : v) {
+            cout << (u ? "Yes\n" : "No\n");
+        }
+    }
+    return 0;
+} 

search-with-noob/peak_element.cpp

@@ -0,0 +1,79 @@
+// https://practice.geeksforgeeks.org/problems/peak-element/1/
+// An element is called a peak element if its value is not smaller than the value of its adjacent elements(if they exists).
+// Given an array arr[] of size N, find the index of any one of its peak elements.
+
+//Binary Search is used
+
+#include<stdio.h>
+#include<stdbool.h>
+
+int peakElement(int a[], int n)
+{
+   int l=0, u=n-1, mid;
+   while(l<=u)
+   {
+       mid= l+(u-l)/2;
+       if(mid==n-1 && a[mid]>=a[mid-1])
+        return mid;
+       else if(mid==0 && a[mid]>=a[mid+1])
+        return mid;
+       else if(a[mid]>=a[mid-1] && a[mid]>=a[mid+1])
+        return mid;
+       else 
+         l=mid+1;
+   }
+   l=0, u=n-1, mid;
+   while(l<=u)
+   {
+       mid= l+(u-l)/2;
+       if(mid==n-1 && a[mid]>=a[mid-1])
+        return mid;
+       else if(mid==0 && a[mid]>=a[mid+1])
+        return mid;
+       else if(a[mid]>=a[mid-1] && a[mid]>=a[mid+1])
+        return mid;
+       else 
+        return u=mid-1;
+   }
+   return -1;
+}
+
+int main() {
+	int t;
+	scanf("%d", &t);
+	while(t--)
+	{
+		int n;
+		scanf("%d", &n);
+		int a[n], tmp[n];
+		for(int i=0;i<n;i++)
+		{
+			scanf("%d", &a[i]);
+			tmp[i] = a[i];
+		}
+		bool f=0;
+
+		int A = peakElement(tmp,n);
+
+		if(A<0 && A>=n)
+		    printf("0\n");
+		else
+		{
+		if(n==1 && A==0)
+		f=1;
+		else
+		if(A==0 && a[0]>=a[1])
+		f=1;
+		else if(A==n-1 && a[n-1]>=a[n-2])
+		f=1;
+		else if(a[A]>=a[A+1] && a[A]>= a[A-1])
+		f=1;
+		else
+		f=0;
+		printf("%d\n", f);
+		}
+
+	}
+
+	return 0;
+}