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Commit 31e2104

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JainMaster
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‎5.More on Loops.ipynb

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Binary Pattern"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Print the following pattern for the given number of rows.<br>\n",
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"<br>\n",
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"Pattern for N = 4<br>\n",
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"1111<br>\n",
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"000<br>\n",
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"11<br>\n",
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"0<br>\n",
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"Input format : N (Total no. of rows)<br>\n",
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"<br>\n",
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"Output format : Pattern in N lines<br>\n",
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"<br>\n",
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"Sample Input :<br>\n",
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"5<br>\n",
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"Sample Output :<br>\n",
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"11111<br>\n",
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"0000<br>\n",
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"111<br>\n",
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"00<br>\n",
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"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": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"n=int(input())\n",
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"for i in range(1,n+1):\n",
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" for l in range(n+2-2*i,0,-1):\n",
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" print(1,end=\"\")\n",
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" print()\n",
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" for j in range (n+1-2*i,0,-1):\n",
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" print(0,end=\"\")\n",
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" print()"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Print Number Pyramid"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Print the following pattern for a given n.<br>\n",
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"<br>\n",
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"For eg. N = 6<br>\n",
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"123456<br>\n",
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" 23456<br>\n",
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" 3456<br>\n",
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" 456<br>\n",
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" 56<br>\n",
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" 6<br>\n",
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" 56<br>\n",
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" 456<br>\n",
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" 3456<br>\n",
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" 23456<br>\n",
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"123456<br>\n",
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"<br>\n",
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"Sample Input 1 :<br>\n",
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"4<br>\n",
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"Sample Output 1 :<br>\n",
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"1234<br>\n",
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" 234<br>\n",
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" 34<br>\n",
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" 4<br>\n",
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" 34<br>\n",
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" 234<br>\n",
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"1234"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Diamond of Stars"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Print the following pattern for the given number of rows.\n",
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"<br>Note: N is always odd.\n",
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"<br><b>Pattern for N = 5<b>\n",
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"![Diamond of Star](https://imgur.com/download/MPB9rTl)\n",
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"</b>The dots represent spaces.</b>"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Input format :<br>\n",
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"N (Total no. of rows and can only be odd)<br>\n",
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"<br>\n",
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"Output format :<br>\n",
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"Pattern in N lines<br>\n",
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"<br>\n",
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"Constraints :<br>\n",
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"1 <= N <= 49<br>\n",
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"<br>"
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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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"n=int(input())\n",
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"for i in range(1,n+1):\n",
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" for j in range(1,i):\n",
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" print(\" \",end=\"\")\n",
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" for l in range(i,n+1):\n",
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" print(l,end=\"\")\n",
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" print()\n",
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"for a in range(1,n):\n",
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" for s in range(n-1-a,0,-1):\n",
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" print(\" \",end=\"\")\n",
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" for k in range(n-a,n+1):\n",
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" print(k,end=\"\")\n",
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" print()"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Rectangular numbers"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Print the following pattern for the given number of rows.<br>\n",
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"<br>\n",
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"Pattern for N = 4<br>\n",
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"4444444<br>\n",
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"4333334<br>\n",
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"4322234<br>\n",
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"4321234<br>\n",
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"4322234<br>\n",
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"4333334 <br>\n",
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"4444444<br>\n",
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"<br>\n",
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"Input format : N (Total no. of rows)<br>\n",
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"Output format : Pattern in N lines<br>\n",
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"<br>\n",
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"Sample Input :<br>\n",
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"3<br>\n",
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"Sample Output :<br>\n",
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"33333<br>\n",
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"32223<br>\n",
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"32123<br>\n",
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"32223<br>\n",
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"33333"
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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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"5\n",
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"555555555\n",
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"544444445\n",
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"543333345\n",
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"543222345\n",
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"543212345\n",
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"543222345\n",
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"543333345\n",
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"544444445\n",
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"555555555\n"
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]
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}
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],
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"source": [
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"n=int(input())\n",
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"pattern_1=n\n",
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"\n",
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"for i in range (1,n+1):\n",
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" pattern_3 = n\n",
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" for k in range(1,i):\n",
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" print(pattern_3,end=\"\")\n",
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" pattern_3-=1\n",
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" for j in range (1,2*n+2-2*i):\n",
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" print(pattern_1,end=\"\")\n",
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" pattern_1 -= 1\n",
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"\n",
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" for l in range (1,i):\n",
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" pattern_3+=1\n",
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" print(pattern_3,end=\"\")\n",
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" print(\"\")\n",
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"\n",
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"\n",
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"pattern_1=2\n",
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"\n",
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"for i in range (n,1,-1):\n",
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" pattern_3 = n\n",
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" for k in range(1,i):\n",
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" print(pattern_3,end=\"\")\n",
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" pattern_3-=1\n",
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" for j in range (1,2*n+2-2*i):\n",
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" print(pattern_1,end=\"\")\n",
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" pattern_1+= 1\n",
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"\n",
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" for l in range (1,i):\n",
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" pattern_3+=1\n",
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" print(pattern_3,end=\"\")\n",
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" print(\"\")"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Print the pattern"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Print the following pattern for the given number of rows.\n",
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"<br><b>Pattern for N = 5</b>\n",
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"![PrintThePattern](https://imgur.com/download/bl7P8Wq)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Input format : N (Total no. of rows)<br>\n",
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"<br>\n",
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"Output format : Pattern in N lines<br>\n",
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"<br>\n",
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"Sample Input :<br>\n",
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"4<br>\n",
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"Sample Output :<br>\n",
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"1 2 3 4<br>\n",
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"9 10 11 12<br>\n",
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"13 14 15 16<br>\n",
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"5 6 7 8"
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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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"name": "stdout",
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"output_type": "stream",
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"text": [
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"4\n",
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"1 2 3 4 \n",
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"9 10 11 12 \n",
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"13 14 15 16 \n",
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"5 6 7 8 \n"
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]
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}
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],
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"source": [
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"n=int(input())\n",
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"startvalue=1\n",
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"for i in range(1,n+1):\n",
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" for j in range(startvalue,startvalue+n):\n",
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" print(j,end=\" \")\n",
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" print()\n",
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" if i==(n+1)//2:\n",
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" if n%2!=0:\n",
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" startvalue=n*(n-2)+1\n",
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" else:\n",
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" startvalue=n*(n-1)+1\n",
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" elif i>(n+1)//2:\n",
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" startvalue=startvalue-2*n\n",
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" else:\n",
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" startvalue=startvalue+2*n"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.6"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 4
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}

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