#include <iostream>
#include <vector>
#include <set>

using namespace std;

int getSmallestArea(vector<vector<int>>& grid) {
    int rows = grid.size();
    if (rows == 0) return 0;
    int cols = grid[0].size();
    if (cols == 0) return 0;

    set<int> rowsSet, colsSet;

    for (int i = 0; i < rows; ++i) {
        for (int j = 0; j < cols; ++j) {
            if (grid[i][j] == 1) {
                rowsSet.insert(i);
                colsSet.insert(j);
            }
        }
    }

    int width = colsSet.empty() ? 0 : *colsSet.rbegin() - *colsSet.begin() + 1;
    int height = rowsSet.empty() ? 0 : *rowsSet.rbegin() - *rowsSet.begin() + 1;

    return width * height;
} 

shipping space
Salesforce

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import heapq
def reduce_sum(lst):
    heapq.heapify(lst)
    s = 0
    while len(lst) > 1:
        first = heapq.heappop(lst)
        second = heapq.heappop(lst)
        s += first + second
        heapq.heappush(lst, first + second)
    return s

Reduce the Array
Salesforce

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#include <iostream>
#include <vector>
#include <algorithm>

int getPotentialOfWinner(std::vector<int>& potential, long long k) {
    int n = potential.size();

    int x = potential[0];
    int m = 0;
    for (int i = 1; i < n; i++) {
        if (m != k) {
            if (x > potential[i]) {
                m++;
            } else {
                x = potential[i];
                m = 1;
            }
        }
    }

    return x;
}

int main() {
    std::vector<int> potentials = {3, 2, 1, 4};
    long long k = 2;
    std::cout << getPotentialOfWinner(potentials, k) << std::endl;
    return 0;
}

Potential winner code

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def processExecution(power, minPower, maxPower):
    result = []
    for min_p, max_p in zip(minPower, maxPower):
        count = sum(1 for p in power if min_p <= p <= max_p)
        power_sum = sum(p for p in power if min_p <= p <= max_p)
        result.append((count, power_sum))
    return result

Amazon

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TCS FREE NQT - Biggest Mass Hiring

Graduation Year: 2024

Eligibility: BTech / BE / MTech / ME / MCA / MSc / MS

Experience: Freshers

Salary:
Ninja - 3.36 LPA
Digital - 7 LPA
Prime - 9 LPA for UG and 11.5 LPA for PG

Apply now:-  www.allcoding1.com


Registration End Date: 10 April 2024
Test Date: 26th April Onwards

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🎯Indian Army recruitment for Engineers - male unmarried and below 27 years of age.

Batch : 2024, 2023, 2022 and previous passout batches.

Position : Lieutenant ( Pay Range : 56k-1.77L)

Apply Now:-
https://www.allcoding1.com/2024/04/indian-army.html

🎯Aera Hiring Automation Engineer Intern

Graduation Year: 2023 / 2024

Eligibility: Having a degree in Computer Science, Information Technology, or a related field; graduated in 2023 or after

Location: Pune

Apply Now:- www.allcoding1.com

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🎯Qualcomm Hiring Engineering Interns


Eligibility: Bachelor's or Master's Degree in Electrical Engineering, Computer Science Engineering, Communication Engineering, Electronics & Communications Engineering

1. Software Engineering Intern:
Graduation Year: 2024

2. Hardware Engineering Intern:
Graduation Year: 2025
Have Knowledge in PLL, LNA, OpAmp, CMOS, ADC/DAC, Cadence, SpectreRF, or Layout is required in RF/Analog/Mixed Signal IC Design

Location: Telangana, Bangalore, Chennai, Noida

Apply :- www.allcoding1.com

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🎯Agoda SDE Tech Internship 2024

Graduation Year: 2025 / 2026

Eligibility: Ongoing pursuit of a Bachelor’s or Master’s Degree in Computer Science or a related field.

Internship period: July – Dec 2024

Interview process:
Registration closes: 21st April 2024
Hacker rank test starts: 26th April 2024
In-office interview date: 3rd May 2024

Location: Gurugram, Haryana

Apply Link: https://careersatagoda.com/job/5417820-tech-internship-2024-india-based-gurgaon-office/

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#include <iostream>
#include <string>
using namespace std;

int smallest_possible_number(const string& S) {
    if (S.size() == 1 && S[0] == '0') {
        return -1;
    }

    if (S.find('1') == string::npos) {
        return -1;
    }

    int number = 1;
    int n = S.size();
    for (int i = n - 1; i >= 0; --i) {
        if (S[i] == '1') {
            if (number % (i + 1) != 0) {
                number *= (i + 1);
            }
        } else {
            while (number % (i + 1) == 0) {
                number += 1;
            }
        }
    }
    return number;
}
Base Conversion

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#include<bits/stdc++.h>
using namespace std;

int main(){
    string a,b,c; cin>>a>>b>>c;
    vector<int>cnt(26,0);
    for(auto ele: a){
        cnt[ele-'A']++;
    }
    for(auto ele: b){
        cnt[ele-'A']++;
    }
    for(auto ele: c){
        cnt[ele-'A']--;
    }
    bool flag=true;
    for(auto ele: cnt){
        if(ele<0) flag=false;
    }
    if(flag) cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
}

Magnetic Letters

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private static double[] calculateMovingAverage(int[] array, int K) {
        double[] smoothedArray = new double[array.length - K + 1];
        double sum = 0;

      
        for (int i = 0; i < K; i++) {
            sum += array[i];
        }

      
        for (int i = 0; i <= array.length - K; i++) {
            if (i > 0) {
              
                sum = sum - array[i - 1] + array[i + K - 1];
            }
            smoothedArray[i] = sum / K;
        }

        return smoothedArray;
    }

Moving Averages

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Moving Average
Python 3

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Money Count

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Case Changer

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#include <iostream>
#include <vector>
using namespace std;

vector<int> HeightProblem(int n, vector<int>& arr) {
    vector<int> ans(n);

    for (int i = 0; i < n; ++i) {
        int height = arr[i];
        int closestLeftHeight = -1;

        for (int j = i - 1; j >= 0; --j) {
            if (arr[j] < height) {
                closestLeftHeight = arr[j];
                break;
            }
        }

        ans[i] = closestLeftHeight;
    }

    return ans;
}

Height problem

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#include<bits/stdc++.h>
#define ll long long
using namespace std;

ll solve(vector<int>a)
{
   set<int>s(a.begin(),a.end());
   ll mex=0;
    while (s.count(mex)) mex++;
   ll ans=distance(s.lower_bound(mex),s.end());
   if(ans==a.size()) return -2;
   return ans;
}
int main() {
    int n; cin>>n;
    vector<int>arr(n);
    for(int i=0;i<n;i++) cin>>arr[i];
    cout<<solve(arr);
    return 0;
}

Mex Number

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