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keshavsingh4522keshavsingh4522
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SUDO PLACEMENtT 2019
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‎GeeksForGeeks/sudo placement 2019.ipynb

+253-8
Original file line numberDiff line numberDiff line change
@@ -251,16 +251,261 @@
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"for _ in range(int(input())):\n",
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" n=int(input())\n",
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" a=list(map(int,input().split()))\n",
254-
" c=a.copy()\n",
255-
" for i in range(len(a)-1):\n",
256-
" for j in range(i+1,len(a)):\n",
257-
" if a[i]<a[j]:\n",
258-
" c.remove(a[i])\n",
259-
" break\n",
260-
" for i in c:\n",
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" print(i,end=' ')\n",
254+
"# c=a.copy()\n",
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"# for i in range(len(a)-1):\n",
256+
"# for j in range(i+1,len(a)):\n",
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"# if a[i]<a[j]:\n",
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"# c.remove(a[i])\n",
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"# break\n",
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" c=[a[-1]]\n",
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" for i in range(len(a)-2,-1,-1):\n",
262+
" if a[i]>= c[-1]:\n",
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" c.append(a[i])\n",
264+
"# for i in c:\n",
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"# print(i,end=' ')\n",
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" print(*(c[::-1]),end='')\n",
262267
" print()"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"1\n",
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"4\n",
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"1 2 3 4\n",
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"4 3 2 1\n"
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]
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}
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],
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"source": [
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"# Reverse an array ---> Infosys, Moonfrog Labs\n",
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"# Difficulty: School   Marks: 0\n",
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"'''\n",
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"Given an array A of size N, print the reverse of it.\n",
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"\n",
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"Input:\n",
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"First line contains an integer denoting the test cases 'T'. T testcases follow. Each testcase contains two lines of input. First line contains N the size of the array A. The second line contains the elements of the array.\n",
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"\n",
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"Output:\n",
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"For each testcase, in a new line, print the array in reverse order.\n",
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"\n",
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"Constraints:\n",
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"1 <= T <= 100\n",
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"1 <= N <=100\n",
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"0 <= Ai <= 100\n",
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"\n",
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"Example:\n",
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"Input:\n",
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"1\n",
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"4\n",
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"1 2 3 4\n",
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"Output:\n",
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"4 3 2 1\n",
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"'''\n",
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"for _ in range(int(input())):\n",
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" n=int(input())\n",
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" a=list(map(int,input().split()))\n",
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" a.reverse()\n",
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" print(*a)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 26,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"1\n",
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"6 2\n",
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"10 20 30 40 50 60\n",
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"20 10 40 30 60 50 "
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]
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}
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],
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"source": [
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"# Reverse array in groups ---> Amazon\n",
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"# Difficulty: Basic   Marks: 1\n",
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"'''\n",
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"Given an array arr[] of positive integers of size N. Reverse every sub-array of K group elements.\n",
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"\n",
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"Input:\n",
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"The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines of input. The first line of each test case consists of an integer N(size of array) and an integer K separated by a space. The second line of each test case contains N space separated integers denoting the array elements.\n",
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"\n",
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"Output:\n",
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"For each test case, print the modified array.\n",
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"\n",
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"Constraints:\n",
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"1 ≤ T ≤ 200\n",
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"1 ≤ N, K ≤ 107\n",
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"1 ≤ A[i] ≤ 1018\n",
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"\n",
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"Example:\n",
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"Input\n",
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"2\n",
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"5 3\n",
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"1 2 3 4 5\n",
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"6 2\n",
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"10 20 30 40 50 60\n",
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"\n",
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"Output\n",
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"3 2 1 5 4\n",
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"20 10 40 30 60 50\n",
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"\n",
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"Explanation:\n",
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"Testcase 1: Reversing groups in size 3, first group consists of elements 1, 2, 3. Reversing this group, we have elements in order as 3, 2, 1.\n",
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"'''\n",
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"from math import ceil\n",
367+
"for i in range(int(input())):\n",
368+
" n,k=map(int,input().split())\n",
369+
" arr=list(map(int,input().split()))\n",
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" arr[:k]=arr[k-1::-1] #i was having a bit difficulty in converting the first group inside the loop\n",
371+
" for i in range(1,ceil(float(n)/k)):\n",
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" arr[k*i:k*i+k]=arr[k*i+k-1:k*i-1:-1]\n",
373+
" print(*arr)\n",
374+
"'''for _ in range(int(input())):\n",
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" n1,n2=input().split()\n",
376+
" a=list(map(int,input().split()))\n",
377+
"# if int(n1)%int(n2)!=0:\n",
378+
"# n2=int(n1)//int(n2)+1\n",
379+
" j=int(n1)//int(n2)\n",
380+
" for i in range(j):\n",
381+
" c=[]\n",
382+
" for k in range(int(n2)*i,int(n2)*(i+1)):\n",
383+
" c.append(a[k])\n",
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" print(*(c[::-1]),end=' ')\n",
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" print('',*(a[:int(n2)*j-1:-1]),end='',sep=' ')\n",
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" print()\n",
387+
"'''\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 7,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"2\n",
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"6\n",
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"7 10 4 3 20 15\n",
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"3\n",
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"7\n",
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"5\n",
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"7 10 4 20 15\n",
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"4\n",
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"15\n"
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]
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}
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],
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"source": [
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"# Kth smallest element ---> ABCO, Accolite, Amazon, Cisco, Hike, Microsoft, Snapdeal, VMWare\n",
413+
"# Difficulty: Medium   Marks: 4\n",
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"'''\n",
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"Given an array arr[] and a number K where K is smaller than size of array, the task is to find the Kth smallest element in the given array. It is given that all array elements are distinct.\n",
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"\n",
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"Expected Time Complexity: O(n)\n",
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"\n",
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"Input:\n",
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"The first line of input contains an integer T, denoting the number of testcases. Then T test cases follow. Each test case consists of three lines. First line of each testcase contains an integer N denoting size of the array. Second line contains N space separated integer denoting elements of the array. Third line of the test case contains an integer K.\n",
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"\n",
422+
"Output:\n",
423+
"Corresponding to each test case, print the kth smallest element in a new line.\n",
424+
"\n",
425+
"Constraints:\n",
426+
"1 <= T <= 100\n",
427+
"1 <= N <= 105\n",
428+
"1 <= arr[i] <= 105\n",
429+
"1 <= K <= N\n",
430+
"\n",
431+
"Example:\n",
432+
"Input:\n",
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"2\n",
434+
"6\n",
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"7 10 4 3 20 15\n",
436+
"3\n",
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"5\n",
438+
"7 10 4 20 15\n",
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"4\n",
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"\n",
441+
"Output:\n",
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"7\n",
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"15\n",
444+
"\n",
445+
"Explanation:\n",
446+
"Testcase 1: 3rd smallest element in the given array is 7.\n",
447+
"'''\n",
448+
"for _ in range(int(input())):\n",
449+
" n=int(input())\n",
450+
" a=list(map(int,input().split()))\n",
451+
" i=int(input())\n",
452+
" a.sort()\n",
453+
" print(a[i-1])"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"[55, 9, 7, 6, 6, 5, 5, 4, 3, 3, 1, 1]"
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]
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},
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"execution_count": 2,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"a=[1,3,4,5,6,7,3,1,9,6,55,5]\n",
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"a.sort(reverse=True)\n",
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"a"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"1\n",
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"5\n",
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"3 1 2 5 4\n",
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"5 3 1 4 2\n"
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]
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}
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],
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"source": [
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"for _ in range(int(input())):\n",
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" n=int(input())\n",
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" a=list(map(int,input().split()))\n",
498+
" a.sort(reverse=True)\n",
499+
" print(*(a[::2]),end=' ')\n",
500+
" print(*(a[1::2]))"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {

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