Merge branch 'main' into main

Author: Dheeraj-06

Problems/Binary-Search/Day-06/sol/Ishan_Raj_Singh/Solution1.py

@@ -0,0 +1,46 @@
+"""
+Submission Link: https://codeforces.com/contest/2014/submission/356141020
+
+Approach:
+1. If n <= 2, impossible to have more than half unhappy -> return -1
+2. Sort the wealth array
+3. Calculate current sum
+4. Find the median position (n//2)
+5. Use mathematical formula: For median person to be unhappy:
+   (sum + X) / (2 * n) > a[median]
+   Solving: X > 2 * n * a[median] - sum
+   Therefore: X = 2 * n * a[median] - sum + 1
+
+"""
+
+def solve():
+    n = int(input())
+    a = list(map(int, input().split()))
+    
+    if n <= 2:
+        print(-1)
+        return
+    
+    a.sort()
+    
+    total_sum = sum(a)
+    
+    mid = n // 2
+    
+    current_avg = total_sum / n
+    unhappy_count = sum(1 for wealth in a if wealth < current_avg / 2)
+    
+    if unhappy_count > n / 2:
+        print(0)
+        return
+    
+    X = 2 * n * a[mid] - total_sum + 1
+    
+    if X < 0:
+        print(0)
+    else:
+        print(X)
+
+t = int(input())
+for _ in range(t):
+    solve()

Problems/Binary-Search/Day-06/sol/Ishan_Raj_Singh/Solution2.py

@@ -0,0 +1,45 @@
+"""
+Approach:
+For each power r (2 to 64):
+- Calculate k ≈ n^(1/r)
+- Check if 1 + k + k² + ... + k^r = n
+- Verify k-1, k, k+1 for floating point precision
+
+Submission Link: https://codeforces.com/contest/1846/submission/356143831
+"""
+
+def solve():
+    n = int(input())
+    
+    if n < 3:
+        print("NO")
+        return
+    
+    for r in range(2, 65):
+        k = int(n ** (1.0 / r))
+        
+        for candidate_k in [max(2, k - 1), k, k + 1]:
+            if candidate_k < 2:
+                continue
+            
+            total = 1
+            power = 1
+            
+            for i in range(r):
+                power *= candidate_k
+                total += power
+                if total > n:
+                    break
+            
+            if total == n:
+                print("YES")
+                return
+        
+        if k < 2:
+            break
+    
+    print("NO")
+
+t = int(input())
+for _ in range(t):
+    solve()

Problems/Binary-Search/Day-07/sol/Ishan_Raj_Singh/Solution1.py

@@ -0,0 +1,45 @@
+"""
+Approach: Two Pointers approach
+Submission Link: https://codeforces.com/problemset/submission/1538/356144559
+"""
+
+def solve_two_pointers():
+    n, l, r = map(int, input().split())
+    a = list(map(int, input().split()))
+    
+    a.sort()
+    count = 0
+    
+    for i in range(n):
+        min_sum = l - a[i]
+        max_sum = r - a[i]
+        
+        left = i + 1
+        right = n - 1
+        
+        lo, hi = i + 1, n
+        while lo < hi:
+            mid = (lo + hi) // 2
+            if a[mid] < min_sum:
+                lo = mid + 1
+            else:
+                hi = mid
+        left_bound = lo
+        
+        lo, hi = i + 1, n
+        while lo < hi:
+            mid = (lo + hi) // 2
+            if a[mid] <= max_sum:
+                lo = mid + 1
+            else:
+                hi = mid
+        right_bound = lo - 1
+        
+        if left_bound < n and right_bound >= i + 1 and left_bound <= right_bound:
+            count += (right_bound - left_bound + 1)
+    
+    print(count)
+
+t = int(input())
+for _ in range(t):
+    solve_two_pointers()

Problems/Binary-Search/Day-07/sol/vishva/soln1.cpp

@@ -0,0 +1,127 @@
+#include<bits/stdc++.h>
+#include<queue>
+using namespace std;
+typedef unordered_map<int, int> umii;
+typedef unordered_map<long long, long long> umll;
+typedef unordered_map<char, long long> umci;
+typedef vector<pair<int, int>> vpi;
+typedef vector<int> vi;
+typedef long long ll;
+typedef vector<long long> vll;
+typedef unordered_map<int , bool> umib;
+#define sum(v) accumulate(v.begin(), v.end(), 0)
+#define endl '\n'
+#define f0(i, n) for(long long i = 0; i < n; i++)
+#define f1(i, n) for(long long i = 1; i < n; i++)
+#define as(v) sort(v.begin(), v.end())
+#define all(x) (x).begin(), (x).end()
+#define pb push_back
+template<class T> umll frequency(vector<T> &v) {umll freq;for(auto &x:v) freq[x]++; return freq;}
+template<class T> umci S_frequency(vector<T> &v) {umci freq;for(auto &x:v) freq[x]++; return freq;}
+template <class T> void input(vector<T> &v){for(auto &x:v)cin>>x;}
+ll power(ll x, ll y){ ll res = 1; while (y > 0){ if (y & 1) res = (ll)(res*x); y = y>>1; x = (ll)(x*x); } return res; }
+void pvll(const vector<long long> &arr){for(auto it : arr){cout << it << " ";}cout << endl;}
+void pvi(const vector<int> &arr){for(auto it : arr){cout << it << " ";}cout << endl;}
+
+
+// submission link :- 
+//https://codeforces.com/contest/1538/submission/352943894
+// C_Number_of_Pairs.cpp
+
+
+
+
+
+
+void solve(){
+
+//-------------INPUT-------------
+    ll n; 
+    cin >> n;
+    ll l,r;
+    cin >> l >> r;
+    vll v(n); input(v);
+    map<ll,ll>mp1;
+    map<ll,ll>mp2;
+    map<ll,ll>mp;
+    ll count=0;
+    as(v);
+    vll temp;
+    
+    for(ll i=0;i<n;i++)
+    {
+        ll req1 = l-v[i];
+        ll a,b;
+        ll req2 = r-v[i];
+        // apply lowerbound to find req1
+        auto it = lower_bound(temp.begin(),temp.end(),req1);
+
+        // using binary search to find the req1 index
+    
+
+        if( it!=temp.end() && *it>=req1)
+        {
+            a = mp[*it];
+        }
+        else
+        {
+            a =-1;
+        }
+        
+        //cout << a << endl;
+
+        // if got a particular a then try to find b
+
+        if(a!=-1 && !temp.empty())
+        {
+            
+            auto it = lower_bound(temp.begin(),temp.end(),req2);
+            //cout << req2 << " " << mp2[*it] << " " << *it << endl;
+            //cout << mp1[*it];
+            if(it == temp.end() && !temp.empty())
+            {
+                b = mp2[*(it-1)]; 
+                //cout << "j";
+            }
+            else if(*it>req2)
+            {
+                b = mp[*it]-1;
+                //cout << "l";
+            }
+            else{
+                b= mp2[*it];
+                //cout << "lf";
+            }
+            if(b>=a)
+            {
+                // this will be total pairs
+                count = count+b-a+1;
+            }
+        }
+        //cout << "b" << " " << b << endl;
+
+        temp.push_back(v[i]);
+        if(mp[v[i]]==0 && v[i]!=v[0])
+        {
+
+            mp[v[i]] = i;
+        }
+        mp2[v[i]] = i;
+    }
+  
+    cout << count << endl;
+
+
+
+//time complexity o(nlogn)
+//-------------CODE--------------
+    
+
+
+}
+
+
+int main(){
+    int tt; cin >> tt; while(tt--)
+{solve();};
+}

contributers.md

@@ -83,5 +83,6 @@
 | YOGESH M | yogesh4216 | CIT Chennai |
 | Krishna Sikheriya | Krishna200608 | IIIT Allahabad |
 | Sarthak Tewari | strongFingers2 | IIIT Allahabad |
+| Gayatri Duse | Gayatrii4506 | KKWIEER |
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