Modify any element of the array - c

Question

Given an array of integers , you can modify any of a number of arbitrary positive integer , and ultimately makes the entire array is strictly increasing and are positive integers , and asked at least need to change a few numbers
input: 5 1 2 2 3 4
output: 3
and there is what i have tried ,Each number in order to reduce more a ( first number minus one , then the second number minus two ,the third number minus three)
#include <stdio.h>
int Modify_the_array(int B[],int n);
int max(int a,int b);
int main(int argc,char *argv) {
int before_array[]={1,2,3,4,1,2,3,4,5};
int len=sizeof(before_array[0])/sizeof(before_array);
int b;
b=Modify_the_array(before_array,len);
printf("%d\n",b);
return 0;
}
int max(int a,int b){
return a>b?a:b;
}
int Modify_the_array(int B[],int len) {
int i,b=0,n=1;
int maxsofar,tmp,j;
for (i=0;i<len;i++){
B[i]=B[i]-n;
n++;
}
maxsofar=0;
tmp=0;
for(i=0;i<len;i++) {
for (j=i+1;j<len;j++) {
if (B[j]==B[i]&&B[i]>1) {
maxsofar=max(maxsofar,++tmp);
b=len-maxsofar;
}
}
}
return b;
}
somebody recommend there is another solution for this question,more efficently ,can anyone give me some advice,thank in advance

Answer 1

I came across the same problem recently. To make clear:
Problem Statement
You are given a sequence of integers a1,a2,a3.....an. You are free to replace any integer with any other positive integer. How many integers must be replaced to make the resulting sequence strictly increasing?
Input Format
The first line of the test case contains an integer N - the number of entries in the sequence.
The next line contains N space separated integers where the ith integer is ai.
Output Format
Output the minimal number of integers that should be replaced to make the sequence strictly increasing.
Given your input, len = 5, arr = [1 2 2 3 4], after minus index+1, get [0 0 -1 -1 -1].
Ignoring negative elements(these must be changed), compute Longest Increasing Subsequence(nondecreasing for this problem), which is a classic Dynamic Programming problem.
Denote the length of LIS = n(these elements will not be changed). So the final answer(the part doesn't belong to the increasing subsequence and the ignored negative part) is len-n(5-2=3).
We can compute LIS in O(nlogn) time with O(n) space.
int solve(vector<int> &arr) {
int len = arr.size();
for(int i = 0; i < len; i++) {
arr[i] -= i+1;
}
vector<int> lis(len,0);
int n = 0;
for(int i = 0; i < len; i++) {
if(arr[i] >= 0) {
int pos = binarysearchPos(lis,n,arr[i]);
lis[pos] = arr[i];
if(n == pos)
n++;
}
}
return len-n;
}
int binarysearchPos(vector<int> &arr, int n, int target) {
if(n == 0)
return 0;
if(arr[n-1] <= target)
return n;
int low = 0, high = n-1;
while(low < high) {
int mid = (low+high)/2;
if(arr[mid] > target) {
high = mid;
} else {
low = mid+1;
}
}
return low;
}