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‎Arrays/AddOneToNumber.cpp

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/*
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Given a non-negative number represented as an array of digits,
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add 1 to the number ( increment the number represented by the digits ).
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The digits are stored such that the most significant digit is at the head of the list.
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Example:
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If the vector has [1, 2, 3]
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the returned vector should be [1, 2, 4]
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as 123 + 1 = 124.
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LINK: https://www.interviewbit.com/problems/add-one-to-number/
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*/
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vector<int> Solution::plusOne(vector<int> &A)
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{
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int c=1;
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for(int i=A.size()-1;i>=0;i--)
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{
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int num = A[i]+c;
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c = num/10;
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A[i] = num%10;
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}
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vector<int> v;
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if(c==1)
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{
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v.push_back(1);
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for(int i=0;i<A.size();i++)
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v.push_back(A[i]);
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}
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else
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{
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int i=0;
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while(i<A.size())
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{
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if(A[i]==0)
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i++;
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else
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break;
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}
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for(;i<A.size();i++)
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v.push_back(A[i]);
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}
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return v;
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}

‎Arrays/AntiDiagonals.cpp

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/*
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Give a N*N square matrix, return an array of its anti-diagonals. Look at the example for more details.
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Example:
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Input:
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1 2 3
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4 5 6
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7 8 9
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Return the following :
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[
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[1],
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[2, 4],
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[3, 5, 7],
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[6, 8],
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[9]
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]
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Input :
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1 2
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3 4
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Return the following :
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[
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[1],
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[2, 3],
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[4]
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]
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LINK: https://www.interviewbit.com/problems/anti-diagonals/
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*/
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vector<vector<int> > Solution::diagonal(vector<vector<int> > &A)
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{
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vector<vector<int>> v;
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int n = A.size();
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for(int i=0;i<n;i++)
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{
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vector<int> vv;
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for(int j=i,k=0;k<=i;j--,k++)
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vv.push_back(A[k][j]);
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v.push_back(vv);
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}
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for(int i=1;i<n;i++)
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{
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vector<int> vv;
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for(int j=n-1,k=i;k<n;j--,k++)
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vv.push_back(A[k][j]);
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v.push_back(vv);
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}
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return v;
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}

‎Arrays/FindDuplicateinArray.cpp

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/*
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Given a read only array of n + 1 integers between 1 and n, find one number that repeats in linear time using less than O(n) space and traversing the stream sequentially O(1) times.
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Sample Input:
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[3 4 1 4 1]
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Sample Output:
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1
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If there are multiple possible answers ( like in the sample case above ), output any one.
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If there is no duplicate, output -1
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LINK: https://www.interviewbit.com/problems/find-duplicate-in-array/
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*/
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int Solution::repeatedNumber(const vector<int> &A) {
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// Do not write main() function.
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// Do not read input, instead use the arguments to the function.
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// Do not print the output, instead return values as specified
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// Still have a doubt. Checkout www.interviewbit.com/pages/sample_codes/ for more details
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int valueRange = A.size()-1;
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int range = sqrt(valueRange);
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if(range*range < valueRange)
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range++;
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int count[range+1];
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memset(count, 0, sizeof(count));
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for(int i=0;i<A.size();i++)
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{
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count[(A[i]-1)/range]++;
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}
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int repeatedRange = -1;
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int numRanges = ((valueRange-1)/range)+1;
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for(int i=0;i<numRanges && repeatedRange==-1;i++)
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{
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if(i<numRanges-1 || valueRange%range==0)
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{
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if(count[i]>range)
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repeatedRange = i;
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}
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else
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if(count[i] > valueRange%range)
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repeatedRange = i;
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}
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if(repeatedRange == -1)
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return -1;
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memset(count, 0, sizeof(count));
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for(int i=0;i<A.size();i++)
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{
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if((A[i]-1)/range == repeatedRange)
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count[(A[i]-1)%range]++;
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}
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for(int i=0;i<range;i++)
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{
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if(count[i]>1)
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return repeatedRange*range+i+1;
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}
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return -1;
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}

‎Arrays/FindPermutation.cpp

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/*
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Given a positive integer n and a string s consisting only of letters D or I, you have to find any permutation of first n positive integer that satisfy the given input string.
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D means the next number is smaller, while I means the next number is greater.
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Notes
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Length of given string s will always equal to n - 1
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Your solution should run in linear time and space.
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Example :
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Input 1:
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n = 3
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s = ID
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Return: [1, 3, 2]
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LINK: https://www.interviewbit.com/problems/find-permutation/
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*/
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vector<int> Solution::findPerm(const string s, int n)
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{
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vector<int> v(n);
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int len = s.length();
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int k=1;
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for(int i=0;i<len;i++)
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{
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if(s[i]=='I')
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{
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v[i] = k;
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k++;
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}
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}
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v[n-1] = k++;
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for(int i=len-1;i>=0;i--)
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{
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if(s[i]=='D')
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{
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v[i] = k;
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k++;
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}
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}
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return v;
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}

‎Arrays/FirstMissingInteger.cpp

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/*
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Given an unsorted integer array, find the first missing positive integer.
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Example:
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Given [1,2,0] return 3,
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[3,4,-1,1] return 2,
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[-8, -7, -6] returns 1
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Your algorithm should run in O(n) time and use constant space.
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LINK: https://www.interviewbit.com/problems/first-missing-integer/
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*/
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int Solution::firstMissingPositive(vector<int> &A)
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{
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int n = A.size();
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int j=0;
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for(int i=0;i<n;i++)
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{
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if(A[i]<=0)
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{
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swap(A[i],A[j]);
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j++;
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}
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}
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int size=n-j;
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for(int i=j;i<n;i++)
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{
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if(abs(A[i])-1+j < n && A[abs(A[i])-1+j]>0)
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A[abs(A[i])-1+j] = -A[abs(A[i])-1+j];
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}
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for(int i=j;i<n;i++)
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if(A[i]>0)
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return (i-j+1);
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return (n-j+1);
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}

‎Arrays/Flip.cpp

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/*
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You are given a binary string(i.e. with characters 0 and 1) S consisting of characters S1, S2, …, SN. In a single operation, you can choose two indices L and R such that 1 ≤ L ≤ R ≤ N and flip the characters SL, SL+1, …, SR. By flipping, we mean change character 0 to 1 and vice-versa.
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Your aim is to perform ATMOST one operation such that in final string number of 1s is maximised. If you don’t want to perform the operation, return an empty array. Else, return an array consisting of two elements denoting L and R. If there are multiple solutions, return the lexicographically smallest pair of L and R.
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Notes:
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Pair (a, b) is lexicographically smaller than pair (c, d) if a < c or, if a == c and b < d.
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For example,
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S = 010
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Pair of [L, R] | Final string
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_______________|_____________
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[1 1] | 110
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[1 2] | 100
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[1 3] | 101
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[2 2] | 000
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[2 3] | 001
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We see that two pairs [1, 1] and [1, 3] give same number of 1s in final string. So, we return [1, 1].
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Another example,
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If S = 111
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No operation can give us more than three 1s in final string. So, we return empty array [].
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LINK: https://www.interviewbit.com/problems/flip/
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*/
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vector<int> Solution::flip(string A)
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{
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vector<int> v;
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int sum=0, msum=0, st=0, end=0;
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int rs = -1, re = -1;
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for(int i=0;i<A.length();i++)
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{
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if(A[i]=='0')
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sum++;
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else
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sum--;
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if(sum<0)
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{
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sum=0;
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st = i+1;
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}
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if(sum>msum)
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{
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msum = sum;
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rs = st;
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re = end = i;
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}
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}
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if(rs!=-1 && re!=-1)
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{
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v.push_back(rs+1);
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v.push_back(re+1);
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}
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return v;
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}

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