Added my solutions to some problems

Bad Triangle/Neha Jhawar/Solution.cpp

@@ -0,0 +1,23 @@
+/**
+Since the array elements are already sorted in ascending order, so sum of 1st 2 elements has to be greater than last element for valid solutions.
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+	int te;
+	cin>>te;
+	while(te--) {
+	    int n, i;
+	    cin>>n;
+	    vector<long> v(n);
+	    for(i=0; i<n; i++) {
+	        cin>>v[i];
+	    }
+	    if(v[0] + v[1] > v[n-1]) cout<<-1<<endl;
+	    else cout<<1<<" "<<2<<" "<<n<<endl;
+	}
+
+	return 0;
+}

Chef and Recipe/Neha Jhawar/ChefAndRecipeSolution.cpp

@@ -0,0 +1,67 @@
+/**
+In this problem we have to keep track of 2 things for a possible solution
+1. Check if no a particular set of number(ingredients) is appearing together
+2. The count of numbers in each set must be unique
+
+So to do that I am using three maps in my solution.
+First map 'mp' keeps track of last position of occurence of the number in the vector v
+Second map 'count' stores the count of each numbers in the vector vector
+Third map 'counter' checks whether a particular element is not repeated in map 'count'
+
+#Please note instead of array A, I am using vector v. 
+*/
+
+
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+	int te;
+	cin>>te;
+	while(te--) {
+	    int n, i;
+	    cin>>n;
+	    vector<int>v(n);
+	    for(i=0; i<n; i++) {
+	        cin>>v[i];
+	    }
+	    map<int, int> mp, count;
+	    int flag = 0;
+	    for(i=0; i<n; i++) {
+	        count[v[i]]++;
+			//checks if a particular element is present in map 'mp'
+	        if(mp.find(v[i])!=mp.end()) {
+				//if the element is present then the difference of value of map corresponding to key and the current element position(i) should be 1
+	            if(i-mp[v[i]]!=1) {
+	                flag=1;
+	                break;
+	            }
+	            mp[v[i]] = i;
+	        }
+	        else {
+	            mp[v[i]]=i;
+	        }
+
+	    }
+	    if(flag) cout<<"NO"<<endl;
+	    else {
+	        flag = 0;
+	        map<int, int> counter;
+	        map<int, int>::iterator itr;
+	        for(itr=count.begin(); itr!=count.end(); itr++) {
+	            if(counter.find(itr->second) != counter.end()) {
+    	            flag=1;
+    	            break;
+    	        }
+    	        else {
+    	            counter[itr->second]=1;
+    	        }
+	        }
+	        if(flag) cout<<"NO"<<endl;
+	        else cout<<"YES"<<endl;
+	    }
+	}
+
+	return 0;
+}

Chef and Work/Neha Jhawar/ChefAndWork.cpp

@@ -0,0 +1,34 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+	int te;
+	cin>>te;
+	while(te--) {
+	    int n, k, i;
+	    cin>>n>>k;
+	    vector<int>v(n);
+	    for(i=0; i<n; i++) {
+	        cin>>v[i];
+	    }
+	    int s=0, count=1, flag=0;
+	    for(i=0; i<n; i++) {
+			//Checks if any element in the vector is greater than k. If yes then loop will be terminated
+	        if(v[i]>k) {
+	            flag=1;
+	            break;
+	        }
+			//counting the current weight the Chef is carrying. If the weight is greater than k then the Chef will drop the item and counter will be incremented by 1.
+	        s += v[i];
+	        if(s>k) {
+	            i--;
+	            s=0;
+	            count++;
+	        }
+	    }
+	    if(flag) cout<<-1<<endl;
+	    else cout<<count<<endl;
+	}
+
+	return 0;
+}

Dynamic Array/Neha Jhawar/DynamicArraySolution.cpp

@@ -0,0 +1,43 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+vector<int> dynamicArray(int n, vector<vector<int>> q) {
+    vector<vector<int>> v(n);
+    vector<int>ar;
+    int i, la=0, sum;
+    for(i=0; i<q.size(); i++) {
+        int bla=la, a, b;
+        if(q[i][0]==1) {
+            a = (q[i][1]^la)%n;
+            v[a].push_back(q[i][2]);
+        }
+        else {
+            a = (q[i][1]^la)%n;
+            b = q[i][2]%v[a].size();
+            la = v[a][b];
+            ar.push_back(la);
+        }
+    }
+    return ar;
+}
+
+int main()
+{
+    int n, q;
+    cin>>n>>q;
+
+    vector<vector<int>> queries(q, vector<int>(3));
+
+    for (int i = 0; i < q; i++) {
+        for (int j = 0; j < 3; j++) {
+            cin>>queries[i][j];
+        }
+    }
+
+    vector<int> result = dynamicArray(n, queries);
+
+    for (int i = 0; i < result.size(); i++) {
+        cout << result[i]<<endl;   
+    }
+    return 0;
+}

Left Rotation/Neha Jhawar/LeftRotationSolution.cpp

@@ -0,0 +1,21 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int main() {
+    int n,d;
+    cin>>n>>d;
+    //d = d%n;
+    int a[n],i;
+    for(i=0; i<n; i++) {
+        cin>>a[i];
+    }
+    for(i=d; i<n; i++)
+        cout<<a[i]<<" ";
+
+    for(i=0; i<d; i++)
+        cout<<a[i]<<" ";
+
+	cout<<endl;
+    return 0;
+}

Omkar and Password/Neha Jhawar/Solution.cpp

@@ -0,0 +1,25 @@
+/**
+If all the numbers in the array are equal then the length of shortest password is length of array else answer is always 1.
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+	int te;
+	cin>>te;
+	while(te--) {
+	    int n, c=0, i;
+	    cin>>n;
+	    vector<int> v(n);
+	    for(i=0; i<n; i++) {
+	        cin>>v[i];
+	    }
+	    for(i=1; i<n; i++) {
+	        if(v[i]==v[0]) c++;
+	    }
+	    if(c==n-1) cout<<n<<endl;
+	    else cout<<1<<endl;
+	}
+	return 0;
+}

Stone/Neha Jhawar/Solution.cpp

@@ -0,0 +1,28 @@
+/**
+First take one stone from the second heap and two stones from the third heap. Then take take one stone from the first heap and two stones from the second heap. Keep track of total count of stones and print it. 
+*/
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+	int te;
+	cin>>te;
+	while(te--) {
+	    int a, b, c, i;
+	    cin>>a>>b>>c;
+	    int count=0;
+	    while(c>=2 && b>0) {
+	        count+=3;
+	        c -= 2;
+	        b--;
+	    }
+	    while(b>=2 && a>0) {
+	        count+=3;
+	        b -= 2;
+	        a--;
+	    }
+	    cout<<count<<endl;
+	}
+
+	return 0;
+}