Question: Objectives Segmented Least Squares Running time Segmented Least Squares DP[0] = 0; for (i = 1; i error then (*) leastErr = error; pointer[i] =
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Objectives Segmented Least Squares Running time Segmented Least Squares DP[0] = 0; for (i = 1; i error then (*) leastErr = error; pointer[i] = j; end end DPl least Err; end Analyse the running time of the above algorithm as a function of n - the number of observations. Assuming that Line (*) requires c*(-i) primitive operations for some constant c
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